JEE Hyperbola Questions

We are given the hyperbola $$H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with one focus at $$(\sqrt{10}, 0)$$ and the corresponding directrix $$x = \frac{9}{\sqrt{10}}$$.

For a hyperbola, the focus is at $$(ae, 0)$$ and the directrix is $$x = \frac{a}{e}$$. So we have:

$$ae = \sqrt{10} \quad \text{...(1)}$$

$$\frac{a}{e} = \frac{9}{\sqrt{10}} \quad \text{...(2)}$$

Multiplying equations (1) and (2): $$ae \cdot \frac{a}{e} = \sqrt{10} \cdot \frac{9}{\sqrt{10}}$$, which gives $$a^2 = 9$$, so $$a = 3$$.

From equation (1): $$3e = \sqrt{10}$$, so $$e = \frac{\sqrt{10}}{3}$$.

Therefore $$e^2 = \frac{10}{9}$$.

Now, $$b^2 = a^2(e^2 - 1) = 9\left(\frac{10}{9} - 1\right) = 9 \cdot \frac{1}{9} = 1$$.

The length of the latus rectum is $$l = \frac{2b^2}{a} = \frac{2 \cdot 1}{3} = \frac{2}{3}$$.

Finally, $$9(e^2 + l) = 9\left(\frac{10}{9} + \frac{2}{3}\right) = 9\left(\frac{10}{9} + \frac{6}{9}\right) = 9 \cdot \frac{16}{9} = 16$$.

Hence, the correct answer is Option C.

The standard form of a rectangular-axis hyperbola with centre at the origin and transverse axis along the x-axis is $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$$.

The foci are $$(\pm c,0)$$ where $$c^{2}=a^{2}+b^{2} \; -(1)$$.

The given point $$P(4,\,2\sqrt{3})$$ lies on the hyperbola. Substituting in the equation of the curve,

$$\frac{4^{2}}{a^{2}}-\frac{(2\sqrt3)^{2}}{b^{2}} = 1 \;\Longrightarrow\; \frac{16}{a^{2}}-\frac{12}{b^{2}} = 1 \; -(2)$$

Let the focal distances of $$P$$ be $$PF_{1}$$ and $$PF_{2}$$. Because the foci are $$(\pm c,0)$$,

$$PF_{1}^{\,2} = (4-c)^{2} + (2\sqrt3)^{2} = (4-c)^{2} + 12 = c^{2}-8c+28 \; -(3)$$
$$PF_{2}^{\,2} = (4+c)^{2} + 12 = c^{2}+8c+28 \; -(4)$$

The product of the focal distances is given to be 32:

$$PF_{1}\,PF_{2}=32 \;\Longrightarrow\; PF_{1}^{\,2}\,PF_{2}^{\,2}=32^{2}=1024$$

Using $$(3)$$ and $$(4)$$:

$$\bigl(c^{2}-8c+28\bigr)\bigl(c^{2}+8c+28\bigr)=1024$$

Employ $$(A-B)(A+B)=A^{2}-B^{2}$$ with $$A=c^{2}+28,\;B=8c$$:

$$\bigl(c^{2}+28\bigr)^{2}-(8c)^{2}=1024$$
$$c^{4}+56c^{2}+784-64c^{2}=1024$$
$$c^{4}-8c^{2}-240=0$$

Put $$z=c^{2}$$ to obtain $$z^{2}-8z-240=0$$. Solving the quadratic,

$$z = \frac{8\pm\sqrt{8^{2}+4\cdot240}}{2} = \frac{8\pm32}{2}$$

The positive root is $$z=20$$, so

$$c^{2}=20,\qquad c = 2\sqrt5 \; -(5)$$

From $$(1)$$, $$a^{2}+b^{2}=20 \; -(6)$$. From $$(2)$$, $$\dfrac{16}{a^{2}}-\dfrac{12}{b^{2}}=1 \; -(7)$$.

Let $$a^{2}=A,\;b^{2}=B$$. Then $$(6)$$ gives $$A+B=20 \; -(8)$$, and $$(7)$$ gives $$\dfrac{16}{A}-\dfrac{12}{B}=1 \; -(9)$$.

Re-write $$(9)$$: $$\dfrac{16}{A} = 1+\dfrac{12}{B} = \dfrac{B+12}{B}$$,
so $$A = \frac{16B}{B+12} \; -(10)$$.

Substitute $$(10)$$ into $$(8)$$:

$$\frac{16B}{B+12}+B = 20$$
$$16B + B(B+12) = 20(B+12)$$
$$B^{2}+28B = 20B + 240$$
$$B^{2}+8B-240 = 0$$

Solving, $$B = \frac{-8\pm\sqrt{8^{2}+4\cdot240}}{2} = \frac{-8\pm32}{2}$$.

Select the positive value: $$B = 12 \;\Longrightarrow\; b^{2}=12$$.

Using $$(8)$$, $$a^{2}=20-12 = 8$$, hence $$a = \sqrt8 = 2\sqrt2$$.

Conjugate axis length
The conjugate axis has length $$p = 2b = 2\sqrt{b^{2}} = 2\sqrt{12}=4\sqrt3$$,
so $$p^{2} = (4\sqrt3)^{2}=48$$.

Latus rectum length
For the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$, the length of a latus rectum is $$q = \frac{2b^{2}}{a}$$.

$$q = \frac{2\cdot12}{2\sqrt2} = \frac{24}{2\sqrt2}= \frac{12}{\sqrt2}=6\sqrt2$$,
so $$q^{2} = (6\sqrt2)^{2}=72$$.

Finally,

$$p^{2}+q^{2}=48+72=120$$.

Hence, the required value is $$\boxed{120}$$.

The standard form of a hyperbola with centre at the origin and transverse axis along the x-axis is
$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \qquad\left(a\gt 0,\;b\gt 0\right).$$

Its foci are $$\left(\pm c,0\right)$$ where $$c^{2}=a^{2}+b^{2}$$.

One focus is given as $$P(-3,0)$$, hence
$$c = 3 \quad\text{and}\quad c^{2}=9.$$ Therefore
$$a^{2}+b^{2}=9 \;.(1)$$

The other focus is $$S(3,0)$$. Through this focus the latus-rectum is the line $$x=3$$ whose end-points are $$Q\!\left(3,\;\frac{b^{2}}{a}\right), \quad R\!\left(3,\;-\frac{b^{2}}{a}\right).$$ The segment $$QR$$ subtends a right angle at the point $$P(-3,0)$$, i.e.
$$\angle QPR = 90^{\circ}.$$

For perpendicular vectors their dot product is zero. Vector $$\overrightarrow{PQ} = (3-(-3),\,\tfrac{b^{2}}{a}-0) = (6,\tfrac{b^{2}}{a})$$
Vector $$\overrightarrow{PR} = (3-(-3),\,-\tfrac{b^{2}}{a}-0) = (6,\,-\tfrac{b^{2}}{a}).$$

Dot product condition:
$$\overrightarrow{PQ}\!\cdot\!\overrightarrow{PR}=0 \;\Longrightarrow\; 6\cdot6+\frac{b^{2}}{a}\left(-\frac{b^{2}}{a}\right)=0 \;\Longrightarrow\; 36-\frac{b^{4}}{a^{2}}=0.$$ Hence
$$b^{4}=36a^{2}\;\Longrightarrow\; b^{2}=6a \;.(2)$$

Substitute $$b^{2}=6a$$ from $$(2)$$ into $$(1)$$:
$$a^{2}+6a-9=0.$$ Solving the quadratic,
$$a=\frac{-6\pm\sqrt{36+36}}{2}=\frac{-6\pm6\sqrt{2}}{2} =-3\pm3\sqrt{2}.$$ Since $$a\gt0,$$ take
$$a = -3+3\sqrt{2}=3(\sqrt{2}-1).$$

Now
$$b^{2}=6a=6\!\left[3(\sqrt{2}-1)\right]=18(\sqrt{2}-1).$$

The product required is
$$a^{2}b^{2} =\Bigl[\,3(\sqrt{2}-1)\Bigr]^{2}\,\,\Bigl[\,18(\sqrt{2}-1)\Bigr]$$ $$=9(3-2\sqrt{2})\times18(\sqrt{2}-1)$$ $$=162\,(3-2\sqrt{2})(\sqrt{2}-1).$$

Compute the remaining product:
$$(3-2\sqrt{2})(\sqrt{2}-1)=5\sqrt{2}-7.$$ Thus
$$a^{2}b^{2}=162\,(5\sqrt{2}-7)=810\sqrt{2}-1134.$$ Hence $$\alpha = 810,\quad \beta = 1134,\quad a^{2}b^{2}= \alpha\sqrt{2}-\beta.$$

The required sum is $$\alpha+\beta = 810+1134 = 1944.$$

Therefore, the value of $$\alpha+\beta$$ is $$1944$$.

The standard horizontal hyperbola with centre at the origin is written as $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$.
Its basic facts are:

• Foci $$\left(\pm c,\,0\right)$$ where $$c^{2}=a^{2}+b^{2}$$.
• Eccentricity $$e=\frac{c}{a}\ (e\gt 1)$$.
• Directrices $$x=\pm\frac{a}{e}$$ (each focus has the nearer directrix).

We are told that one focus is $$(-5,0)$$ and the corresponding directrix is $$5x+9=0$$, i.e. $$x=-\frac{9}{5}$$.
Hence

$$c=5,\qquad -\frac{a}{e}=-\frac{9}{5}\; \Longrightarrow\; \frac{a}{e}=\frac{9}{5}\; -(1)$$

Using $$e=\frac{c}{a}$$, substitute $$c=5$$ into $$e=\frac{5}{a}$$ and then into $$(1)$$:

$$\frac{a}{e}=a\left(\frac{a}{5}\right)=\frac{a^{2}}{5}=\frac{9}{5}\; \Longrightarrow\; a^{2}=9\; \Longrightarrow\; a=3.$$

Now compute $$b^{2}$$ using $$c^{2}=a^{2}+b^{2}$$:

$$25=9+b^{2}\; \Longrightarrow\; b^{2}=16\; \Longrightarrow\; b=4.$$

Therefore the hyperbola is $$\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$$.

The given point $$\left(\alpha,\,2\sqrt{5}\right)$$ lies on the curve, so substitute it:

$$\frac{\alpha^{2}}{9}-\frac{(2\sqrt{5})^{2}}{16}=1 \quad\Longrightarrow\quad \frac{\alpha^{2}}{9}-\frac{20}{16}=1.$$

Simplify:

$$\frac{\alpha^{2}}{9}-\frac{5}{4}=1\; \Longrightarrow\; \frac{\alpha^{2}}{9}=\frac{9}{4}\; \Longrightarrow\; \alpha^{2}=\frac{81}{4}\; \Longrightarrow\; \alpha=\pm\frac{9}{2}.$$

Let $$P(\alpha,2\sqrt{5})$$ be the point; its distances to the two foci $$F_{1}(5,0)$$ and $$F_{2}(-5,0)$$ are

$$PF_{1}=\sqrt{(\alpha-5)^{2}+(2\sqrt{5})^{2}},\qquad PF_{2}=\sqrt{(\alpha+5)^{2}+(2\sqrt{5})^{2}}.$$

The required product is $$p=PF_{1}\,PF_{2}=\sqrt{\bigl[(\alpha-5)^{2}+20\bigr]\bigl[(\alpha+5)^{2}+20\bigr]}.$$

$$\alpha=\frac{9}{2}=4.5$$

$$\bigl[(\alpha-5)^{2}+20\bigr]=(4.5-5)^{2}+20=0.25+20=20.25,$$ $$\bigl[(\alpha+5)^{2}+20\bigr]=(4.5+5)^{2}+20=90.25+20=110.25.$$

Product inside the root: $$20.25\times110.25=2232.5625,$$ hence

$$p=\sqrt{2232.5625}=47.25.$$

$$\alpha=-\frac{9}{2}=-4.5$$

This merely interchanges the two factors, giving the same product $$p=47.25$$.

Finally, $$4p=4\times47.25=189.$$

Hence, $$\boxed{4p=189}$$.

The two foci are $$(4,2)$$ and $$(8,2)$$, so the transverse axis is horizontal and its midpoint is the centre of the hyperbola.

Centre $$\,(h,k)=\left(\frac{4+8}{2},\,2\right)=(6,2)$$.

Let the hyperbola in standard form be$$(x-h)^2/a^2-\,(y-k)^2/b^2=1.$$ Here $$c$$ is the focal distance from the centre, given by $$c=\lvert8-6\rvert=2\quad\Rightarrow\quad c^2=4.$$ For a rectangular hyperbola,$$c^2=a^2+b^2\quad-(1).$$

After shifting the origin to $$(6,2)$$ and clearing denominators we obtain $$b^2\,(x-6)^2-a^2\,(y-2)^2=a^2b^2\quad-(2).$$ Expanding $$(2)$$ gives $$b^2(x^2-12x+36)-a^2(y^2-4y+4)=a^2b^2.$$ Collect the terms on the left: $$b^2x^2-12b^2x+36b^2-a^2y^2+4a^2y-4a^2-a^2b^2=0.$$ The required general form is $$3x^2-y^2-\alpha x+\beta y+\gamma=0,$$ so we match coefficients.

Coefficient of $$x^2:$$ $$b^2=3.$$ Coefficient of $$y^2:$$ $$-a^2=-1\;\Rightarrow\;a^2=1.$$ With $$a^2=1,\;b^2=3,$$ equation $$-(1)$$ gives $$c^2=a^2+b^2=1+3=4,$$ agreeing with $$c=2,$$ so the values are consistent.

Substituting $$a^2=1,\;b^2=3$$ in $$b^2(x-6)^2-a^2(y-2)^2=a^2b^2$$: $$3(x-6)^2-(y-2)^2=3.$$ Expand: $$3(x^2-12x+36)-(y^2-4y+4)=3,$$ $$3x^2-36x+108-y^2+4y-4=3.$$ Move everything to one side: $$3x^2-y^2-36x+4y+108-4-3=0,$$ $$3x^2-y^2-36x+4y+101=0.$$ Therefore $$-\alpha=-36\;\Rightarrow\;\alpha=36,$$ $$\beta=4,$$ $$\gamma=101.$$

Finally, $$\alpha+\beta+\gamma=36+4+101=141.$$

The required value is $$141.$$

We are given two hyperbolas $$H_1: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ and $$H_2: -\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$$ with the latus rectum of $$H_1$$ equal to $$15\sqrt{2}$$, the latus rectum of $$H_2$$ equal to $$12\sqrt{5}$$, the eccentricity $$e_1 = \sqrt{\frac{5}{2}}$$, and the product of their transverse axes equal to $$100\sqrt{10}$$. Our goal is to find $$25e_2^2$$.

For the first hyperbola, the transverse axis is $$2a$$, the latus rectum is $$\frac{2b^2}{a}$$, and we have $$b^2 = a^2(e_1^2 - 1)$$. For the second hyperbola (in its conjugate form), the transverse axis is $$2B$$, the latus rectum is $$\frac{2A^2}{B}$$, and $$A^2 = B^2(e_2^2 - 1)$$.

Since $$e_1 = \sqrt{\frac{5}{2}}$$, it follows that $$e_1^2 = \frac{5}{2}$$ and hence $$b^2 = a^2\left(\frac{5}{2} - 1\right) = \frac{3a^2}{2}$$. Substituting into the expression for the latus rectum of $$H_1$$ gives $$\frac{2b^2}{a} = \frac{2\cdot\frac{3a^2}{2}}{a} = 3a = 15\sqrt{2}$$, which yields $$a = 5\sqrt{2}$$.

Therefore the transverse axis of $$H_1$$ is $$2a = 10\sqrt{2}$$. Since the product of the transverse axes is $$(2a)(2B) = 100\sqrt{10}$$, we have $$10\sqrt{2}\cdot 2B = 100\sqrt{10}$$. This gives $$2B = \frac{100\sqrt{10}}{10\sqrt{2}} = 10\sqrt{5}$$ and hence $$B = 5\sqrt{5}$$.

Turning to the second hyperbola, its latus rectum satisfies $$\frac{2A^2}{B} = 12\sqrt{5}$$, so $$A^2 = \frac{12\sqrt{5}\cdot B}{2} = \frac{12\sqrt{5}\cdot 5\sqrt{5}}{2} = \frac{300}{2} = 150$$.

Using the relation $$A^2 = B^2(e_2^2 - 1)$$ gives $$150 = 125(e_2^2 - 1)$$, so $$e_2^2 - 1 = \frac{150}{125} = \frac{6}{5}$$ and therefore $$e_2^2 = 1 + \frac{6}{5} = \frac{11}{5}$$.

Finally, we compute $$25e_2^2 = 25 \times \frac{11}{5} = 55$$.

The answer is 55

Consider the ellipse given by $$\frac{x^2}{9} + \frac{y^2}{25} = 1.$$ 

Since the larger denominator is 25, the major axis runs along the y-axis, so we set $$a^2 = 25$$ and $$b^2 = 9.$$ 

 $$c^2 = a^2 - b^2 = 25 - 9 = 16 \implies c = 4,$$ and the foci lie at $$(0,\pm4).$$

The eccentricity of this ellipse is $$e_1 = \frac{c}{a} = \frac{4}{5}.$$

foci at $$(0,\pm4)$$, so its transverse axis is also vertical and its focal parameter is $$c_h = 4.$$ 

 $$e_2 = \frac{15}{8}\,e_1 = \frac{15}{8}\times\frac{4}{5} = \frac{3}{2}.$$ Thus $$a_h = \frac{c_h}{e_2} = \frac{4}{3/2} = \frac{8}{3},$$ and $$b_h^2 = c_h^2 - a_h^2 = 16 - \frac{64}{9} = \frac{80}{9}.$$ 

the equation of the hyperbola is $$\frac{y^2}{64/9} \;-\;\frac{x^2}{80/9} \;=\;1.$$

$$P\bigl(\sqrt{2},\,\tfrac{14}{3}\sqrt{\tfrac{2}{5}}\bigr)$$ lies on this curve, note first that $$y^2 = \frac{196}{9}\times\frac{2}{5} = \frac{392}{45},$$ so $$\frac{y^2}{64/9} = \frac{392/45}{64/9} = \frac{392}{45}\times\frac{9}{64} = \frac{49}{40},$$ while $$\frac{x^2}{80/9} = \frac{2}{80/9} = \frac{9}{40}.$$ Their difference is $$\frac{49}{40} - \frac{9}{40} = 1,$$ confirming that $$P$$ indeed satisfies the hyperbola’s equation.

For any point on a hyperbola with eccentricity $$e_2$$ and transverse semi-axis $$a_h$$, the distances to the two foci satisfy $$|PF| = \bigl|e_2\,y \pm a_h\bigr|.$$

Since the y-coordinate of $$P$$ is positive, we compute $$e_2\,y = \frac{3}{2}\times\frac{14}{3}\sqrt{\frac{2}{5}} = 7\sqrt{\frac{2}{5}},$$

hence the two focal distances are $$PF_1 = 7\sqrt{\frac{2}{5}} - \frac{8}{3},\qquad PF_2 = 7\sqrt{\frac{2}{5}} + \frac{8}{3}.$$ The smaller of these is $$7\sqrt{\frac{2}{5}} - \frac{8}{3}.$$

Therefore, the smaller focal distance is $$7\sqrt{\frac{2}{5}} - \frac{8}{3}$$.

Ellipse: $$\frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$$. $$a^2=100, b^2=75$$. $$c^2=25$$, $$c=5$$. $$e_E = 1/2$$. Foci: $$(1\pm5, 1) = (6,1)$$ and $$(-4,1)$$.

Hyperbola eccentricity $$e_H = 1/e_E = 2$$. Same foci, centre at (1,1).

For hyperbola: $$c_H = 5$$ (same foci), $$e_H = 2$$, so $$a_H = c_H/e_H = 5/2$$.

$$b_H^2 = c_H^2 - a_H^2 = 25-25/4 = 75/4$$.

Transverse axis $$\alpha = 2a_H = 5$$, conjugate axis $$\beta = 2b_H = 2\sqrt{75/4} = \sqrt{75} = 5\sqrt{3}$$.

$$3\alpha^2 + 2\beta^2 = 3(25) + 2(75) = 75+150 = 225$$.

The correct answer is Option 4: 225.

A hyperbola $$H$$ centered at the origin with foci on the x-axis is given, and a circle $$C_1$$ centered at the origin touches $$H$$ and has area $$36\pi$$ while another circle $$C_2$$ centered at one focus touches $$H$$ at its vertex and has area $$4\pi$$.

From the areas we get $$r_1^2 = 36 \implies r_1 = 6$$ and $$r_2^2 = 4 \implies r_2 = 2$$.

The circle $$C_1$$ is tangent to the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ at its closest point to the origin, namely the vertex $$(a,0)$$, so $$r_1 = a = 6$$.

Since the circle $$C_2$$ is centered at the focus $$(ae,0)$$ and touches the hyperbola at the vertex $$(a,0)$$, its radius is the distance from the focus to the vertex: $$r_2 = ae - a = a(e - 1) = 6(e - 1) = 2 \implies e - 1 = \frac{1}{3} \implies e = \frac{4}{3}$$.

Then $$b^2 = a^2(e^2 - 1) = 36\left(\frac{16}{9} - 1\right) = 36 \times \frac{7}{9} = 28$$.

Finally, the length of the latus rectum is $$\ell = \frac{2b^2}{a} = \frac{2 \times 28}{6} = \frac{56}{6} = \frac{28}{3}$$, so the correct answer is Option B: $$\frac{28}{3}$$.

Hyperbola $$H: -\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, eccentricity $$e = \sqrt{3}$$, latus rectum $$\frac{2a^2}{b} = 4\sqrt{3}$$.

$$e^2 = 1 + \frac{a^2}{b^2} \implies 3 = 1 + \frac{a^2}{b^2} \implies a^2 = 2b^2$$.

From latus rectum, $$\frac{2(2b^2)}{b} = 4\sqrt{3} \implies 4b = 4\sqrt{3} \implies b = \sqrt{3} \implies b^2 = 3$$. Then $$a^2 = 6$$.

Point $$(\alpha, 6)$$ lies on $$H \implies -\frac{\alpha^2}{6} + \frac{36}{3} = 1 \implies \frac{\alpha^2}{6} = 11 \implies \alpha^2 = 66$$.

Product of focal distances for a vertical hyperbola is $$\beta = (ey_1)^2 - b^2 = (3)(36) - 3 = 108 - 3 = 105$$.

Value: $$\alpha^2 + \beta = 66 + 105 = \mathbf{171}$$

Hyperbola: $$\frac{x^2}{9} - \frac{y^2}{4} = 1$$. So $$a = 3, b = 2, c = \sqrt{13}$$. Foci: $$(\pm\sqrt{13}, 0)$$.

Let $$P = (x_0, y_0)$$ in first quadrant on hyperbola.

Area of triangle with foci $$F_1(-\sqrt{13},0)$$ and $$F_2(\sqrt{13},0)$$:

Area = $$\frac{1}{2} \times 2\sqrt{13} \times y_0 = \sqrt{13} \cdot y_0 = 2\sqrt{13}$$.

So $$y_0 = 2$$. From hyperbola: $$\frac{x_0^2}{9} - \frac{4}{4} = 1 \Rightarrow \frac{x_0^2}{9} = 2 \Rightarrow x_0^2 = 18$$.

Distance from origin squared: $$x_0^2 + y_0^2 = 18 + 4 = 22$$.

The answer is Option (3): $$\boxed{22}$$.

We are given a hyperbola with eccentricity $$e$$, latus rectum length 9, and directrices $$x = \pm \dfrac{4}{\sqrt{13}}$$.

For a standard hyperbola $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$ the directrices are $$x = \pm \dfrac{a}{e}$$ and the length of the latus rectum is $$\dfrac{2b^2}{a}$$.

From the directrices $$\dfrac{a}{e} = \dfrac{4}{\sqrt{13}}$$ so $$a = \dfrac{4e}{\sqrt{13}}$$, and from the latus rectum $$\dfrac{2b^2}{a} = 9$$ so $$b^2 = \dfrac{9a}{2}$$. Using $$b^2 = a^2(e^2 - 1)$$ we obtain

$$\dfrac{9a}{2} = a^2(e^2 - 1)$$

$$\dfrac{9}{2} = a(e^2 - 1)$$

Substituting $$a = \dfrac{4e}{\sqrt{13}}$$ gives

$$\dfrac{9}{2} = \dfrac{4e}{\sqrt{13}}(e^2 - 1)$$

$$\dfrac{9\sqrt{13}}{8} = e(e^2 - 1) = e^3 - e$$

If we try $$e = \dfrac{\sqrt{13}}{2}$$ then

$$e^3 - e = \dfrac{13\sqrt{13}}{8} - \dfrac{\sqrt{13}}{2} = \dfrac{13\sqrt{13} - 4\sqrt{13}}{8} = \dfrac{9\sqrt{13}}{8}$$

This works, so $$e = \dfrac{\sqrt{13}}{2}$$. Hence $$a = \dfrac{4 \cdot \frac{\sqrt{13}}{2}}{\sqrt{13}} = 2$$ and $$b^2 = a^2(e^2 - 1) = 4\left(\dfrac{13}{4} - 1\right) = 9$$. Thus the equation of the hyperbola is $$\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$$.

The tangent line is given by $$y - \sqrt{3}x + \sqrt{3} = 0$$, or $$y = \sqrt{3}(x - 1)$$. Its slope is $$m_t = \sqrt{3}$$. The tangent to $$\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$$ at a point $$(x_0,y_0)$$ has equation

$$\dfrac{x\,x_0}{4} - \dfrac{y\,y_0}{9} = 1$$

or equivalently $$y = \dfrac{9x_0}{4y_0}x - \dfrac{9}{y_0}$$. Comparing with $$y = \sqrt{3}x - \sqrt{3}$$ gives $$\dfrac{9x_0}{4y_0} = \sqrt{3}$$ and $$-\dfrac{9}{y_0} = -\sqrt{3}$$, so $$y_0 = 3\sqrt{3}$$. Then $$x_0 = \dfrac{4\sqrt{3}\,y_0}{9} = 4$$. Hence the point of tangency is $$(4,3\sqrt{3})$$, which indeed lies on the hyperbola since $$\dfrac{16}{4} - \dfrac{27}{9} = 1$$.

The foci are at $$(\pm ae,0) = (\pm\sqrt{13},0)$$. For the point $$(x_0,y_0)$$ on the right branch the focal distances are $$d_1 = ex_0 - a = 2\sqrt{13}-2$$ and $$d_2 = ex_0 + a = 2\sqrt{13}+2$$. Their product is

$$m = d_1\,d_2 = (2\sqrt{13})^2 - 4 = 52 - 4 = 48$$.

Finally,

$$4e^2 + m = 4 \cdot \dfrac{13}{4} + 48 = 13 + 48 = 61$$

Therefore the required value is $$\boxed{61}$$.

For the hyperbola $$\frac{x^2}{3} - \frac{y^2}{5} = 1$$: $$a^2 = 3, b^2 = 5$$, so $$c^2 = a^2 + b^2 = 8$$, $$c = 2\sqrt{2}$$.

Focus on positive x-axis: $$S = (2\sqrt{2}, 0)$$.

Circle C has centre $$A(\sqrt{6}, \sqrt{5})$$ and passes through S. Radius = $$|AS|$$.

$$|AS|^2 = (2\sqrt{2}-\sqrt{6})^2 + (0-\sqrt{5})^2 = (8 - 4\sqrt{12} + 6) + 5 = 19 - 4\sqrt{12}$$

$$= 19 - 8\sqrt{3}$$

SAB is a diameter, so B is the diametrically opposite point of S through centre A.

$$B = 2A - S = (2\sqrt{6} - 2\sqrt{2}, 2\sqrt{5})$$.

Now find the area of triangle OSB, where $$O = (0,0)$$, $$S = (2\sqrt{2}, 0)$$, $$B = (2\sqrt{6}-2\sqrt{2}, 2\sqrt{5})$$.

Area = $$\frac{1}{2}|x_S \cdot y_B - x_B \cdot y_S|$$

$$= \frac{1}{2}|2\sqrt{2} \cdot 2\sqrt{5} - (2\sqrt{6}-2\sqrt{2}) \cdot 0|$$

$$= \frac{1}{2} \cdot 4\sqrt{10} = 2\sqrt{10}$$

Square of area = $$(2\sqrt{10})^2 = 40$$.

The answer is $$\boxed{40}$$.

For the hyperbola $$\frac{x^2}{9} - \frac{y^2}{b^2} = 1$$, we have $$a = 3$$.

The semi-latus rectum is $$l = \frac{b^2}{a} = \frac{b^2}{3}$$. The end of the latus rectum is at $$(ae, \frac{b^2}{a})$$, where $$c = ae = \sqrt{9 + b^2}$$.

The latus rectum subtends angle $$\frac{\pi}{3}$$ at the centre. The half-angle at centre is $$\frac{\pi}{6}$$.

$$\tan\frac{\pi}{6} = \frac{b^2/3}{\sqrt{9+b^2}}$$

$$\frac{1}{\sqrt{3}} = \frac{b^2}{3\sqrt{9+b^2}}$$

$$3\sqrt{9+b^2} = \sqrt{3} b^2$$

Squaring: $$9(9+b^2) = 3b^4$$

$$81 + 9b^2 = 3b^4$$

$$3b^4 - 9b^2 - 81 = 0$$

$$b^4 - 3b^2 - 27 = 0$$

Using the quadratic formula with $$u = b^2$$:

$$u = \frac{3 + \sqrt{9 + 108}}{2} = \frac{3 + \sqrt{117}}{2} = \frac{3 + 3\sqrt{13}}{2}$$

So $$b^2 = \frac{3}{2}(1 + \sqrt{13})$$.

Here $$l = 3, m = 2, n = 13$$ (with $$l, m$$ coprime).

$$l^2 + m^2 + n^2 = 9 + 4 + 169 = 182$$.

Therefore, the answer is $$\boxed{182}$$.

We seek the point $$P(x_0, y_0)$$ on the hyperbola $$3x^2 - 4y^2 = 36$$ that is closest to the line $$3x + 2y = 1$$, and then compute $$\sqrt{2}(y_0 - x_0)\,.$$

First, rewrite the hyperbola in standard form as $$\frac{x^2}{12} - \frac{y^2}{9} = 1\,, $$ so that $$a^2 = 12$$ and $$b^2 = 9\,. $$ A convenient parametrization is for the right branch $$x = 2\sqrt{3}\cosh t,\quad y = 3\sinh t$$ and for the left branch $$x = -2\sqrt{3}\cosh t,\quad y = 3\sinh t\,. $$

The distance from a point $$(x,y)$$ to the line $$3x + 2y - 1 = 0$$ is given by $$d = \frac{\bigl|3x + 2y - 1\bigr|}{\sqrt{13}}\,. $$ On the right branch we set $$f(t) = 6\sqrt{3}\cosh t + 6\sinh t - 1$$ since $$3x + 2y - 1 = 6\sqrt{3}\cosh t + 6\sinh t - 1\,. $$ Differentiating yields $$f'(t) = 6\sqrt{3}\sinh t + 6\cosh t = 0 \quad\Longrightarrow\quad \tanh t = -\frac{1}{\sqrt{3}}\,. $$ Together with $$\cosh^2 t - \sinh^2 t = 1$$ one finds $$\cosh t = \sqrt{\frac{3}{2}},\qquad \sinh t = -\frac{1}{\sqrt{2}}\,. $$ Hence the critical point on the right branch is $$x_0 = 2\sqrt{3}\,\sqrt{\tfrac{3}{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2},\qquad y_0 = 3\Bigl(-\frac{1}{\sqrt{2}}\Bigr) = -\frac{3}{\sqrt{2}} = -\frac{3\sqrt{2}}{2}\,. $$

On the left branch we instead have $$f(t) = -6\sqrt{3}\cosh t + 6\sinh t - 1\,, $$ and setting $$f'(t)=0$$ gives $$\tanh t = \frac{1}{\sqrt{3}}\,, $$ leading to $$x_0 = -3\sqrt{2},\qquad y_0 = \frac{3\sqrt{2}}{2}\,. $$

To decide which point is nearer, compute the numerators of the distances. For the right‐branch point $$\bigl(3\sqrt{2},-\tfrac{3\sqrt{2}}{2}\bigr)$$ one has $$\bigl|3(3\sqrt{2}) + 2\bigl(-\tfrac{3\sqrt{2}}{2}\bigr) - 1\bigr| = \bigl|9\sqrt{2} - 3\sqrt{2} - 1\bigr| = \bigl|6\sqrt{2} - 1\bigr|\approx 7.49\,. $$ For the left‐branch point $$\bigl(-3\sqrt{2},\tfrac{3\sqrt{2}}{2}\bigr)$$ one gets $$\bigl|3(-3\sqrt{2}) + 2\bigl(\tfrac{3\sqrt{2}}{2}\bigr) - 1\bigr| = \bigl|-9\sqrt{2} + 3\sqrt{2} - 1\bigr| = \bigl|-6\sqrt{2} - 1\bigr|\approx 9.49\,. $$ Thus the right branch point is closer to the line.

Finally, at that nearest point we compute $$\sqrt{2}(y_0 - x_0) = \sqrt{2}\Bigl(-\frac{3\sqrt{2}}{2} - 3\sqrt{2}\Bigr) = \sqrt{2}\,\Bigl(-\tfrac{9\sqrt{2}}{2}\Bigr) = -\frac{9\times 2}{2} = -9\,. $$ The correct answer is Option C: $$-9\,. $$

We need to find the area of a region defined by the hyperbola's axes and a parabola.

Rewrite the hyperbola in standard form

$$ 16x^2 - y^2 + 64x + 4y + 44 = 0 $$

$$ 16(x^2 + 4x) - (y^2 - 4y) + 44 = 0 $$

$$ 16(x^2 + 4x + 4) - (y^2 - 4y + 4) + 44 - 64 + 4 = 0 $$

$$ 16(x + 2)^2 - (y - 2)^2 = 16 $$

$$ \frac{(x+2)^2}{1} - \frac{(y-2)^2}{16} = 1 $$

Center: $$(-2, 2)$$, $$a = 1$$, $$b = 4$$.

Identify axes

Transverse axis T: $$y = 2$$ (horizontal line through center)

Conjugate axis C: $$x = -2$$ (vertical line through center)

Set up the region

Region: above parabola $$x^2 = y + 4$$ (i.e., $$y = x^2 - 4$$), below transverse axis $$y = 2$$, and to the right of conjugate axis $$x = -2$$.

Find intersection points

Parabola meets $$y = 2$$: $$x^2 = 6$$, so $$x = \sqrt{6}$$ (taking $$x > -2$$).

Also $$x = -\sqrt{6} \approx -2.449 < -2$$.

Parabola meets $$x = -2$$: $$y = 4 - 4 = 0$$.

So the region is bounded by $$x = -2$$ (left), $$y = 2$$ (top), and $$y = x^2 - 4$$ (bottom), from $$x = -2$$ to $$x = \sqrt{6}$$.

Compute the area

$$ A = \int_{-2}^{\sqrt{6}} [2 - (x^2 - 4)]\,dx = \int_{-2}^{\sqrt{6}} (6 - x^2)\,dx $$

$$ = \left[6x - \frac{x^3}{3}\right]_{-2}^{\sqrt{6}} $$

At $$x = \sqrt{6}$$: $$6\sqrt{6} - \frac{6\sqrt{6}}{3} = 6\sqrt{6} - 2\sqrt{6} = 4\sqrt{6}$$

At $$x = -2$$: $$-12 + \frac{8}{3} = -\frac{28}{3}$$

$$ A = 4\sqrt{6} - \left(-\frac{28}{3}\right) = 4\sqrt{6} + \frac{28}{3} $$

This matches Option 2: $$4\sqrt{6} + \frac{28}{3}$$.

The answer is $$\boxed{4\sqrt{6} + \frac{28}{3}}$$.

Given: Hyperbola $$H$$ with vertices $$(\pm 6, 0)$$ and eccentricity $$e = \frac{\sqrt{5}}{2}$$.

So $$a = 6$$, $$c = ae = 3\sqrt{5}$$, $$b^2 = c^2 - a^2 = 45 - 36 = 9$$.

Hyperbola: $$\frac{x^2}{36} - \frac{y^2}{9} = 1$$.

Find the point on $$H$$ in the first quadrant where the normal has slope $$-\sqrt{2}$$ (parallel to $$\sqrt{2}x + y = 2\sqrt{2}$$).

For the hyperbola, $$y' = \frac{9x}{36y} = \frac{x}{4y}$$. Normal slope $$= -\frac{4y}{x} = -\sqrt{2}$$.

$$y = \frac{x\sqrt{2}}{4}$$

Substituting into the hyperbola equation:

$$\frac{x^2}{36} - \frac{x^2 \cdot 2}{16 \cdot 9} = 1 \implies \frac{x^2}{36} - \frac{x^2}{72} = 1 \implies \frac{x^2}{72} = 1 \implies x^2 = 72$$

$$x = 6\sqrt{2}$$, $$y = \frac{6\sqrt{2} \cdot \sqrt{2}}{4} = 3$$.

Normal at $$(6\sqrt{2}, 3)$$ with slope $$-\sqrt{2}$$:

$$y - 3 = -\sqrt{2}(x - 6\sqrt{2}) \implies y = -\sqrt{2}x + 12 + 3 = -\sqrt{2}x + 15$$

The normal meets the $$y$$-axis at $$(0, 15)$$. The line segment $$N$$ runs from $$(6\sqrt{2}, 3)$$ to $$(0, 15)$$.

$$d^2 = (6\sqrt{2})^2 + (15 - 3)^2 = 72 + 144 = \boxed{216}$$

The tangent to the parabola $$y^2 = 12x$$ at $$(3, \alpha)$$ is perpendicular to $$2x + 2y = 3$$.

Find $$\alpha$$.

Since $$(3, \alpha)$$ is on $$y^2 = 12x$$: $$\alpha^2 = 36 \implies \alpha = \pm 6$$.

Slope of tangent at $$(3, \alpha)$$: differentiating $$y^2 = 12x$$ gives $$\frac{dy}{dx} = \frac{6}{y} = \frac{6}{\alpha}$$.

Slope of line $$2x + 2y = 3$$ is $$-1$$.

For perpendicularity: $$\frac{6}{\alpha} \times (-1) = -1 \implies \alpha = 6$$.

Set up the hyperbola.

With $$\alpha = 6$$: $$36x^2 - 9y^2 = 9 \times 36$$ simplifies to $$\frac{x^2}{9} - \frac{y^2}{36} = 1$$.

Here $$a^2 = 9, b^2 = 36$$.

Verify the point on the hyperbola.

Point $$(\alpha - 1, \alpha + 2) = (5, 8)$$: $$\frac{25}{9} - \frac{64}{36} = \frac{25}{9} - \frac{16}{9} = 1$$ ✓

Find the normal at $$(5, 8)$$.

Slope of tangent: $$\frac{dy}{dx} = \frac{b^2 x}{a^2 y} = \frac{36 \times 5}{9 \times 8} = \frac{5}{2}$$

Slope of normal: $$-\frac{2}{5}$$

Normal line: $$y - 8 = -\frac{2}{5}(x - 5)$$ which gives $$2x + 5y = 50$$.

Distance from $$(6, -4)$$ to the normal.

$$d = \frac{|2(6) + 5(-4) - 50|}{\sqrt{4 + 25}} = \frac{|12 - 20 - 50|}{\sqrt{29}} = \frac{58}{\sqrt{29}} = \frac{2 \times 29}{\sqrt{29}} = 2\sqrt{29}$$

$$d^2 = 4 \times 29 = 116$$

The answer is $$\boxed{116}$$.

We are given the hyperbola $$H: \frac{y^2}{25} - \frac{x^2}{16} = 1$$ and the point $$P(4, 1)$$.

Here $$a^2 = 25$$ and $$b^2 = 16$$.

For a hyperbola $$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$, the tangent line $$y = mx + c$$ satisfies the condition $$c^2 = a^2 - b^2 m^2$$. Let us verify this.

Substituting $$y = mx + c$$ into $$\frac{y^2}{25} - \frac{x^2}{16} = 1$$:

$$\frac{(mx+c)^2}{25} - \frac{x^2}{16} = 1$$

$$16(mx+c)^2 - 25x^2 = 400$$

$$(16m^2 - 25)x^2 + 32mcx + 16c^2 - 400 = 0$$

For tangency, discriminant = 0: $$(32mc)^2 = 4(16m^2 - 25)(16c^2 - 400)$$

Expanding: $$1024m^2c^2 = 1024m^2c^2 - 25600m^2 - 1600c^2 + 40000$$

$$25600m^2 + 1600c^2 = 40000$$

$$16m^2 + c^2 = 25$$

So the tangent condition is $$c^2 = 25 - 16m^2$$.

The tangent through $$P(4,1)$$: since $$1 = 4m + c$$, we get $$c = 1 - 4m$$.

$$(1-4m)^2 = 25 - 16m^2$$

$$1 - 8m + 16m^2 = 25 - 16m^2$$

$$32m^2 - 8m - 24 = 0$$

$$4m^2 - m - 3 = 0$$

$$(4m + 3)(m - 1) = 0$$

So $$m_1 = 1$$ and $$m_2 = -\frac{3}{4}$$.

Therefore $$|m_1| = 1$$ and $$|m_2| = \frac{3}{4}$$.

Now we find the tangent lines with slopes $$|m_1| = 1$$ and $$|m_2| = \frac{3}{4}$$ that make positive x-intercepts.

For slope 1: $$c^2 = 25 - 16 = 9$$, so $$c = \pm 3$$. The x-intercept is $$-\frac{c}{m} = -c$$. For a positive intercept, we need $$c = -3$$. Tangent: $$y = x - 3$$, with $$\alpha = 3$$.

For slope $$\frac{3}{4}$$: $$c^2 = 25 - 16 \cdot \frac{9}{16} = 25 - 9 = 16$$, so $$c = \pm 4$$. The x-intercept is $$-\frac{c}{3/4} = -\frac{4c}{3}$$. For a positive intercept, we need $$c = -4$$. Tangent: $$y = \frac{3}{4}x - 4$$, with $$\beta = \frac{16}{3}$$.

Point $$Q$$ is the intersection of these two tangent lines:

$$x - 3 = \frac{3}{4}x - 4$$

$$\frac{x}{4} = -1 \Rightarrow x = -4, \quad y = -7$$

So $$Q = (-4, -7)$$.

$$(PQ)^2 = (4-(-4))^2 + (1-(-7))^2 = 64 + 64 = 128$$

$$\alpha\beta = 3 \times \frac{16}{3} = 16$$

$$\frac{(PQ)^2}{\alpha\beta} = \frac{128}{16} = 8$$

The answer is $$8$$.

We need to find the smallest even natural number $$n$$ such that the eccentricity of the hyperbola $$H_n: \frac{x^2}{1+n} - \frac{y^2}{3+n} = 1$$ is rational, and then compute $$21l$$ where $$l$$ is the length of the latus rectum.

First, we find the eccentricity in terms of $$n$$.

For a hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, the eccentricity is $$e = \sqrt{1 + \frac{b^2}{a^2}}$$.

Here $$a^2 = 1+n$$ and $$b^2 = 3+n$$, so:

$$ e^2 = 1 + \frac{3+n}{1+n} = \frac{(1+n) + (3+n)}{1+n} = \frac{4+2n}{1+n} $$

Next, we determine when $$e$$ is rational.

Let $$f = n + 1$$. Then $$e^2 = \frac{2f + 2}{f} = 2 + \frac{2}{f}$$.

For $$e$$ to be rational, $$e^2$$ must be the square of a rational number. Let $$e = \frac{p}{q}$$ in lowest terms. Then $$\frac{p^2}{q^2} = 2 + \frac{2}{f}$$, which gives $$f = \frac{2q^2}{p^2 - 2q^2}$$.

For $$f$$ to be a positive integer, we need $$p^2 - 2q^2 > 0$$ and $$p^2 - 2q^2$$ must divide $$2q^2$$.

From this, we search for the smallest even $$n$$.

We need $$n$$ even, so $$f = n + 1$$ must be odd. Testing systematically:

Try $$p = 10, q = 7$$: $$p^2 - 2q^2 = 100 - 98 = 2$$, and $$f = \frac{2 \times 49}{2} = 49$$ (odd). So $$n = 48$$ (even).

We should verify no smaller even $$n$$ works. For smaller values of $$f$$ (odd), we need $$2 + 2/f$$ to be a perfect square of a rational. Checking $$f = 1$$ ($$n=0$$, not natural), $$f = 3$$ ($$n=2$$): $$e^2 = 8/3$$, not a perfect square. $$f = 5$$ ($$n=4$$): $$e^2 = 12/5$$, not rational square. Continuing this way, $$f = 7, 9, 11, \ldots, 47$$ all fail to give a rational $$e$$. Thus $$n = 48$$ is the smallest even value.

Now, we verify $$e$$ is rational for $$n = 48$$.

$$ e^2 = \frac{4 + 96}{49} = \frac{100}{49} \implies e = \frac{10}{7} \quad \text{(rational)} \checkmark $$

Then, we calculate the latus rectum length.

For $$H_{48}$$: $$a^2 = 49$$, $$b^2 = 51$$, $$a = 7$$.

The length of the latus rectum of a hyperbola is $$l = \frac{2b^2}{a}$$:

$$ l = \frac{2 \times 51}{7} = \frac{102}{7} $$

Continuing, we compute $$21l$$.

$$ 21l = 21 \times \frac{102}{7} = 3 \times 102 = 306 $$

The answer is 306.

We are given a hyperbola $$H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with eccentricity $$e = \sqrt{\frac{5}{2}}$$ and latus rectum $$= 6\sqrt{2}$$.

For a hyperbola, $$e^2 = 1 + \frac{b^2}{a^2}$$. Substituting $$\frac{5}{2} = 1 + \frac{b^2}{a^2}$$ gives $$\frac{b^2}{a^2} = \frac{3}{2}$$, so $$b^2 = \frac{3a^2}{2}$$. The length of the latus rectum is $$\frac{2b^2}{a}$$; setting $$\frac{2b^2}{a} = 6\sqrt{2}$$ yields $$\frac{2 \cdot \frac{3a^2}{2}}{a} = 6\sqrt{2}$$ and hence $$3a = 6\sqrt{2}$$, so $$a = 2\sqrt{2}$$. Therefore $$a^2 = 8$$ and $$b^2 = \frac{3 \times 8}{2} = 12$$.

For a line $$y = mx + c$$ to be tangent to the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, the condition is $$c^2 = a^2 m^2 - b^2$$. Substituting $$m = 2$$, $$a^2 = 8$$ and $$b^2 = 12$$ gives $$c^2 = 8 \times 4 - 12 = 32 - 12 = 20$$. Therefore, the correct answer is Option B: $$20$$.

We need to find the point through which the normal to the hyperbola does NOT pass.

The parabola is $$y^2 = 24x$$, so $$4a = 24$$, giving $$a = 6$$.

A parametric point on the parabola is $$(at^2, 2at) = (6t^2, 12t)$$.

Differentiating $$y^2 = 24x$$ implicitly: $$2y \frac{dy}{dx} = 24$$, so $$\frac{dy}{dx} = \frac{12}{y} = \frac{12}{12t} = \frac{1}{t}$$.

The tangent must be perpendicular to $$2x + 2y = 5$$ (slope $$= -1$$), so the tangent slope $$= 1$$.

$$\frac{1}{t} = 1 \implies t = 1$$

So $$(\alpha, \beta) = (6, 12)$$.

Hyperbola: $$\frac{x^2}{36} - \frac{y^2}{144} = 1$$, with $$a^2 = 36$$ and $$b^2 = 144$$.

The point on the hyperbola is $$(\alpha + 4, \beta + 4) = (10, 16)$$.

Verification: $$\frac{100}{36} - \frac{256}{144} = \frac{400 - 256}{144} = \frac{144}{144} = 1$$ ✓

Differentiating $$\frac{x^2}{36} - \frac{y^2}{144} = 1$$ implicitly:

$$\frac{2x}{36} - \frac{2y}{144}\frac{dy}{dx} = 0$$

$$\frac{dy}{dx} = \frac{144x}{36y} = \frac{4x}{y}$$

At $$(10, 16)$$: $$\frac{dy}{dx} = \frac{40}{16} = \frac{5}{2}$$

The slope of the normal is $$m_n = -\frac{1}{dy/dx} = -\frac{2}{5}$$.

The normal at $$(10, 16)$$:

$$y - 16 = -\frac{2}{5}(x - 10)$$

$$5(y - 16) = -2(x - 10)$$

$$5y - 80 = -2x + 20$$

$$2x + 5y = 100$$

Option A: $$(25, 10)$$: $$2(25) + 5(10) = 50 + 50 = 100$$ ✓

Option B: $$(20, 12)$$: $$2(20) + 5(12) = 40 + 60 = 100$$ ✓

Option C: $$(30, 8)$$: $$2(30) + 5(8) = 60 + 40 = 100$$ ✓

Option D: $$(15, 13)$$: $$2(15) + 5(13) = 30 + 65 = 95 \neq 100$$ ✗

Therefore, the correct answer is Option D: $$(15, 13)$$.

We need to find the point through which the normal to the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{9} = 1$$ at $$(8, 3\sqrt{3})$$ passes.

Since $$(8, 3\sqrt{3})$$ lies on the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{9} = 1$$, we have $$\frac{64}{a^2} - \frac{27}{9} = 1 \implies \frac{64}{a^2} - 3 = 1 \implies \frac{64}{a^2} = 4 \implies a^2 = 16$$.

Differentiating $$\frac{x^2}{16} - \frac{y^2}{9} = 1$$ implicitly yields $$\frac{2x}{16} - \frac{2y}{9}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{9x}{16y}$$.

At $$(8, 3\sqrt{3})$$ the slope of the tangent is $$\frac{dy}{dx} = \frac{9 \times 8}{16 \times 3\sqrt{3}} = \frac{72}{48\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$$.

Since the slope of the normal is the negative reciprocal, it is $$-\frac{2}{\sqrt{3}}$$, and its equation through $$(8, 3\sqrt{3})$$ is $$y - 3\sqrt{3} = -\frac{2}{\sqrt{3}}(x - 8)$$.

Substituting $$(-1, 9\sqrt{3})$$ into this equation gives $$9\sqrt{3} - 3\sqrt{3} = -\frac{2}{\sqrt{3}}(-1 - 8)$$, or $$6\sqrt{3} = -\frac{2}{\sqrt{3}} \times (-9) = \frac{18}{\sqrt{3}} = 6\sqrt{3}$$, confirming that $$(-1, 9\sqrt{3})$$ lies on the normal.

Therefore, the required point is $$(-1, 9\sqrt{3})$$.

We need to find which point the hyperbola passes through, given that $$x = 1$$ is a directrix of $$kx^2 - y^2 = 6$$.

$$kx^2 - y^2 = 6 \implies \frac{x^2}{6/k} - \frac{y^2}{6} = 1$$

This is of the form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with $$a^2 = \frac{6}{k}$$ and $$b^2 = 6$$.

$$b^2 = a^2(e^2 - 1)$$

$$6 = \frac{6}{k}(e^2 - 1)$$

$$k = e^2 - 1$$

$$e^2 = k + 1$$

For a hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, the directrices are at $$x = \pm \frac{a}{e}$$.

Given that $$x = 1$$ is a directrix:

$$\frac{a}{e} = 1 \implies a = e$$

$$a^2 = e^2$$

$$\frac{6}{k} = k + 1$$

$$6 = k(k + 1) = k^2 + k$$

$$k^2 + k - 6 = 0$$

$$(k + 3)(k - 2) = 0$$

$$k = 2$$ (taking the positive value since $$k > 0$$ for a valid hyperbola)

$$2x^2 - y^2 = 6$$

Option A: $$(-2\sqrt{5}, 6)$$: $$2(20) - 36 = 40 - 36 = 4 \neq 6$$

Option B: $$(-\sqrt{5}, 3)$$: $$2(5) - 9 = 10 - 9 = 1 \neq 6$$

Option C: $$(\sqrt{5}, -2)$$: $$2(5) - 4 = 10 - 4 = 6 = 6$$ $$\checkmark$$

Option D: $$(2\sqrt{5}, 3\sqrt{6})$$: $$2(20) - 54 = 40 - 54 = -14 \neq 6$$

Therefore, the correct answer is Option C: $$(\sqrt{5}, -2)$$.

The hyperbola is $$a^2x^2 - y^2 = b^2$$.

Rewrite in standard form by dividing by $$b^2$$:

$$ \frac{x^2}{(b/a)^2} - \frac{y^2}{b^2} = 1 $$

This is a hyperbola of the form $$\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$$ where $$A = \frac{b}{a}$$ and $$B = b$$.

The tangent line is $$\lambda x - 2y = \mu$$, i.e., $$y = \frac{\lambda}{2}x - \frac{\mu}{2}$$.

This is of the form $$y = mx + c$$ where $$m = \frac{\lambda}{2}$$ and $$c = -\frac{\mu}{2}$$.

For a line $$y = mx + c$$ to be tangent to $$\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$$, the condition is:

$$ c^2 = A^2m^2 - B^2 $$

Substituting:

$$ \left(-\frac{\mu}{2}\right)^2 = \left(\frac{b}{a}\right)^2\left(\frac{\lambda}{2}\right)^2 - b^2 $$

$$ \frac{\mu^2}{4} = \frac{b^2\lambda^2}{4a^2} - b^2 $$

Multiply through by $$\frac{4}{b^2}$$:

$$ \frac{\mu^2}{b^2} = \frac{\lambda^2}{a^2} - 4 $$

Rearranging:

$$ \frac{\lambda^2}{a^2} - \frac{\mu^2}{b^2} = 4 $$

$$ \left(\frac{\lambda}{a}\right)^2 - \left(\frac{\mu}{b}\right)^2 = 4 $$

The answer is Option D: 4.

For the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$:

Eccentricity: $$e^2 = 1 + \frac{b^2}{a^2}$$, Latus rectum: $$l = \frac{2b^2}{a}$$

For its conjugate hyperbola $$\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$$:

Eccentricity: $$(e')^2 = 1 + \frac{a^2}{b^2}$$, Latus rectum: $$l' = \frac{2a^2}{b}$$

Set up equations from the given conditions:

From $$e^2 = \frac{11}{14}l$$:

$$1 + \frac{b^2}{a^2} = \frac{11}{14} \cdot \frac{2b^2}{a} = \frac{11b^2}{7a}$$

$$\frac{a^2 + b^2}{a^2} = \frac{11b^2}{7a}$$

$$7a(a^2 + b^2) = 11a^2 b^2 \implies 7(a^2 + b^2) = 11ab^2 \quad \cdots (1)$$

From $$(e')^2 = \frac{11}{8}l'$$:

$$1 + \frac{a^2}{b^2} = \frac{11}{8} \cdot \frac{2a^2}{b} = \frac{11a^2}{4b}$$

$$\frac{a^2 + b^2}{b^2} = \frac{11a^2}{4b}$$

$$4b(a^2 + b^2) = 11a^2 b^2 \implies 4(a^2 + b^2) = 11a^2 b \quad \cdots (2)$$

Find the ratio $$b/a$$:

Dividing equation (1) by equation (2):

$$\frac{7}{4} = \frac{11ab^2}{11a^2 b} = \frac{b}{a}$$

Therefore $$b = \frac{7a}{4}$$.

Solve for $$a$$:

Substituting $$b = \frac{7a}{4}$$ into equation (2):

$$4\left(a^2 + \frac{49a^2}{16}\right) = 11a^2 \cdot \frac{7a}{4}$$

$$4 \cdot \frac{16a^2 + 49a^2}{16} = \frac{77a^3}{4}$$

$$\frac{65a^2}{4} = \frac{77a^3}{4}$$

$$65 = 77a \implies a = \frac{65}{77} = \frac{5}{7} \cdot \frac{13}{11}$$

Find $$b$$ and compute $$77a + 44b$$:

$$b = \frac{7}{4} \cdot \frac{65}{77} = \frac{7 \times 65}{4 \times 77} = \frac{65}{44}$$

$$77a + 44b = 77 \times \frac{65}{77} + 44 \times \frac{65}{44} = 65 + 65 = 130$$

The correct answer is Option D: $$130$$.

We need to find the length of the latus rectum of the hyperbola whose foci coincide with those of the ellipse given by $$\dfrac{x^2}{16} + \dfrac{y^2}{7} = 1$$. For this ellipse, $$a^2 = 16$$ and $$b^2 = 7$$, so $$c^2 = a^2 - b^2 = 16 - 7 = 9$$ and hence $$c = 3$$. Therefore, the foci of the ellipse lie at $$(\pm 3,0)\,$$.

Turning to the hyperbola, its equation is $$\dfrac{x^2}{144} - \dfrac{y^2}{\alpha} = \dfrac{1}{25}\,. $$ Dividing both sides by $$\tfrac{1}{25}$$ converts this into the standard form $$\dfrac{x^2}{144/25} - \dfrac{y^2}{\alpha/25} = 1\,, $$ which shows that $$a_h^2 = \dfrac{144}{25}$$ and $$b_h^2 = \dfrac{\alpha}{25}\,$$, so that $$a_h = \dfrac{12}{5}\,.$$

Since the foci of the hyperbola must coincide with those of the ellipse, we set $$c_h = 3$$. For a hyperbola, $$c_h^2 = a_h^2 + b_h^2\,, $$ thus $$c_h^2 = \dfrac{144 + \alpha}{25}$$ and with $$c_h^2 = 9$$ we obtain $$9 = \dfrac{144 + \alpha}{25}\,, $$ which gives $$144 + \alpha = 225$$ and hence $$\alpha = 81\,. $$ It follows that $$b_h^2 = \dfrac{81}{25}\,.$$

The length of the latus rectum of a hyperbola is given by $$\dfrac{2\,b_h^2}{a_h}\,. $$ Substituting the values found above yields

$$\frac{2\,b_h^2}{a_h} = \frac{2 \times \frac{81}{25}}{\frac{12}{5}} = \frac{\frac{162}{25}}{\frac{12}{5}} = \frac{162}{25} \times \frac{5}{12} = \frac{162}{60} = \frac{27}{10}\,.$$

The correct answer is Option D: $$\dfrac{27}{10}\,.$$

The given hyperbola is $$\dfrac{x^{2}}{100}-\dfrac{y^{2}}{64}=1$$.

For this hyperbola  • semi-transverse axis $$a=10$$,   • semi-conjugate axis $$b=8$$,  • eccentricity $$e=\sqrt{1+\dfrac{b^{2}}{a^{2}}}= \sqrt{1+\dfrac{64}{100}}=\dfrac{\sqrt{41}}{5}$$.

Hence the foci are $$S\,(ae,0)=\left(2\sqrt{41},0\right),\qquad S_{1}\,(-ae,0)=\left(-2\sqrt{41},0\right).$$

Let the point $$P(x,y)$$ lie in the first quadrant on the right branch of the hyperbola. Because $$x^{2}/100-y^{2}/64=1$$ and $$x\gt 0,\;y\gt 0$$ we may treat $$x$$ as the only free variable (with $$x\gt 10$$) and write

$$y=\dfrac45\sqrt{x^{2}-100}\qquad\bigl(y\gt 0\bigr).$$

1. Slope of the tangent at P
Differentiate $$\dfrac{x^{2}}{100}-\dfrac{y^{2}}{64}=1:$$ $$\dfrac{2x}{100}-\dfrac{2y}{64}\dfrac{dy}{dx}=0 \;\Longrightarrow\; \dfrac{dy}{dx}= \dfrac{64}{100}\dfrac{x}{y}= \dfrac{16x}{25y}.$$ Hence the tangent slope is $$m=\dfrac{16x}{25y}.$$

2. Line through S having the same slope
Through $$S(2\sqrt{41},0)$$ with slope $$m$$ the line is $$y=m\bigl(x-2\sqrt{41}\bigr) \;\Longrightarrow\; mx-y-2m\sqrt{41}=0.\tag{-1}$$

3. Perpendicular distance of P from this line
Using the distance formula,

$$\delta=\dfrac{\left|mx-y-2m\sqrt{41}\right|}{\sqrt{m^{2}+1}}.$$ Compute the numerator first: $$$ \begin{aligned} mx-y&=\dfrac{16x}{25y}\,x-y =\dfrac{16x^{2}-25y^{2}}{25y}\\ &=\dfrac{16x^{2}-25\Bigl(\dfrac{16}{25}(x^{2}-100)\Bigr)}{25y} =\dfrac{16x^{2}-16x^{2}+1600}{25y} =\dfrac{64}{y}. \end{aligned} $$$ So $$$ mx-y-2m\sqrt{41}= \dfrac{64}{y}- 2\Bigl(\dfrac{16x}{25y}\Bigr)\sqrt{41} =\dfrac{1}{y}\Bigl[64-\dfrac{32x\sqrt{41}}{25}\Bigr]. $$$ Since $$x\gt 10$$ the bracket is negative, and its absolute value is $$$ \left|\;64-\dfrac{32x\sqrt{41}}{25}\right| =\dfrac{32x\sqrt{41}-1600}{25}. $$$ Next, $$$ m^{2}+1= \Bigl(\dfrac{16x}{25y}\Bigr)^{2}+1 =\dfrac{256x^{2}+625y^{2}}{625y^{2}} =\dfrac{16\bigl(41x^{2}-2500\bigr)}{625y^{2}}, $$$ hence $$\sqrt{m^{2}+1}= \dfrac{4\sqrt{41x^{2}-2500}}{25y}.$$ Putting everything together, $$$ \delta=\dfrac{ \dfrac{32x\sqrt{41}-1600}{25y}} { \dfrac{4\sqrt{41x^{2}-2500}}{25y}} =\dfrac{32x\sqrt{41}-1600}{4\sqrt{41x^{2}-2500}} =\dfrac{8x\sqrt{41}-400}{\sqrt{41x^{2}-2500}}.\tag{-2} $$$

4. Distance $$\beta=S_{1}P$$
$$$ \beta^{2}= (x+2\sqrt{41})^{2}+y^{2} =x^{2}+4x\sqrt{41}+164+\dfrac{16}{25}(x^{2}-100) =\dfrac{41}{25}x^{2}+4x\sqrt{41}+100.\tag{-3} $$$

5. The angle $$\alpha=\angle SPS_{1}$$
In triangle $$\triangle SS_{1}P$$ let $$p=SP,\quad q=S_{1}P=\beta,\quad d=SS_{1}=4\sqrt{41}.$$ For a point on the right branch of a hyperbola, $$p-q=2a=20\;\Longrightarrow\; p=q+20.\tag{-4}$$ Using the well-known half-angle identity for any triangle $$\sin\dfrac{\alpha}{2}=\sqrt{\dfrac{(s-p)(s-q)}{pq}},$$ where $$s=\dfrac{p+q+d}{2}$$ is the semiperimeter. Substituting $$p-q=20$$ gives $$$ (s-p)(s-q)=\dfrac{d-20}{2}\cdot\dfrac{d+20}{2} =\dfrac{d^{2}-400}{4} =\dfrac{256}{4}=64, $$$ because $$d^{2}=(4\sqrt{41})^{2}=656.$$ Thus $$$ \sin\dfrac{\alpha}{2}= \sqrt{\dfrac{64}{pq}} =\dfrac{8}{\sqrt{pq}}.\tag{-5} $$$

6. Required expression
$$$ E=\dfrac{\beta\delta}{9}\sin\dfrac{\alpha}{2} =\dfrac{\beta\delta}{9}\cdot\dfrac{8}{\sqrt{p\beta}} =\dfrac{8\delta}{9}\sqrt{\dfrac{\beta}{p}} =\dfrac{8\delta}{9}\sqrt{\dfrac{\beta}{\beta+20}}\quad\bigl(\text{by }( -4)\bigr).\tag{-6} $$$

7. Behaviour of $$E$$ as $$x$$ varies
From $$( -2)$$, $$\delta(x)=\dfrac{8x\sqrt{41}-400}{\sqrt{41x^{2}-2500}}.$$ A simple derivative test (or numerical check) shows that $$\delta(x)$$ is increasing and $$$ \lim_{x\to\infty}\delta(x)=8\quad\text{while}\quad \delta(x)\lt 8\ \text{for every finite }x.  $$$ Likewise, from $$( -3)$$ we have $$\beta(x)\to\infty$$ as $$x\to\infty$$, so $$$ 0\lt \sqrt{\dfrac{\beta}{\beta+20}}\lt 1,\qquad \lim_{x\to\infty}\sqrt{\dfrac{\beta}{\beta+20}}=1. $$$ Therefore, taking the limit in $$( -6)$$, $$$ \boxed{\; \lim_{x\to\infty}E=\dfrac{8}{9}\times 8= \dfrac{64}{9}=7.111\ldots\; }. $$$ For every finite $$x$$ we have $$\delta\lt 8$$ and $$\sqrt{\beta/(\beta+20)}\lt 1$$, hence $$$ E=\dfrac{8\delta}{9}\sqrt{\dfrac{\beta}{\beta+20}} \lt \dfrac{8}{9}\times 8=\dfrac{64}{9}. $$$ Consequently, $$$ 7 \lt E \lt \dfrac{64}{9}\;(=7.111\ldots). $$$ Thus $$E$$ can be made arbitrarily close to $$64/9$$ from below but never reaches or exceeds it. The greatest integer $$\le E$$ for every allowable point $$P$$ is therefore $$7$$.

Answer: 7

We have the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with eccentricity $$e = \frac{5}{4}$$. Since $$e^2 = 1 + \frac{b^2}{a^2}$$ and $$e^2 = \frac{25}{16}$$, it follows that $$\frac{25}{16} = 1 + \frac{b^2}{a^2}$$, which gives $$\frac{b^2}{a^2} = \frac{9}{16}$$.

The point $$\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$$ lies on the hyperbola, so substituting into its equation gives:
$$\frac{64}{5a^2} - \frac{144}{25b^2} = 1$$
Since $$b^2 = \frac{9a^2}{16}$$, this becomes:
$$\frac{64}{5a^2} - \frac{144 \times 16}{25 \times 9a^2} = 1$$
$$\frac{64}{5a^2} - \frac{256}{25a^2} = 1$$
$$\frac{320 - 256}{25a^2} = 1 \implies \frac{64}{25a^2} = 1$$
Therefore $$a^2 = \frac{64}{25}$$ and $$b^2 = \frac{9}{16} \times \frac{64}{25} = \frac{36}{25}$$.

The normal to the hyperbola at $$(x_0,y_0)$$ is given by $$\frac{a^2 x}{x_0} + \frac{b^2 y}{y_0} = a^2 + b^2$$. At $$(x_0,y_0) = \left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$$ this becomes:
$$\frac{(64/25)x}{8/\sqrt{5}} + \frac{(36/25)y}{12/5} = \frac{64}{25} + \frac{36}{25}$$
$$\frac{64\sqrt{5}}{200}x + \frac{36 \times 5}{25 \times 12}y = \frac{100}{25}$$
$$\frac{8\sqrt{5}}{25}x + \frac{3}{5}y = 4$$
Multiplying through by 25 yields $$8\sqrt{5}\,x + 15y = 100$$.

Comparing with the given form $$8\sqrt{5}\,x + \beta y = \lambda$$ shows that $$\beta = 15$$ and $$\lambda = 100$$, hence $$\lambda - \beta = 100 - 15 = 85$$. Therefore, the answer is $$\boxed{85}$$.

Given hyperbola $$H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with $$a > 0, b > 0$$, its transverse axis length is $$2a$$ and its conjugate axis length is $$2b$$, and their sum is $$2a + 2b = 4(2\sqrt{2} + \sqrt{14}),$$ so that $$a + b = 2(2\sqrt{2} + \sqrt{14}) = 4\sqrt{2} + 2\sqrt{14}.$$

The eccentricity is $$e = \frac{\sqrt{11}}{2},$$ hence $$e^2 = \frac{11}{4} = 1 + \frac{b^2}{a^2},$$ giving $$\frac{b^2}{a^2} = \frac{7}{4}, \quad b = \frac{a\sqrt{7}}{2}.$$

Substituting into the sum equation yields $$a + \frac{a\sqrt{7}}{2} = 4\sqrt{2} + 2\sqrt{14},$$ or equivalently $$a\left(1 + \frac{\sqrt{7}}{2}\right) = 4\sqrt{2} + 2\sqrt{14},$$ so that $$a \cdot \frac{2 + \sqrt{7}}{2} = 4\sqrt{2} + 2\sqrt{14}.$$ Noting that $$4\sqrt{2} + 2\sqrt{14} = 2\sqrt{2}(2 + \sqrt{7}),$$ we find $$a = \frac{2 \cdot 2\sqrt{2}(2 + \sqrt{7})}{2 + \sqrt{7}} = 4\sqrt{2}.$$

It follows that $$b = \frac{4\sqrt{2} \cdot \sqrt{7}}{2} = 2\sqrt{14},$$ and hence $$a^2 = 32, \quad b^2 = 4 \times 14 = 56, \quad a^2 + b^2 = 32 + 56 = 88.$$

The answer is $$\boxed{88}$$.

Let $$L_1$$ be a tangent to the hyperbola $$\frac{x^2}{16} - \frac{y^2}{4} = 1$$, and $$L_2$$ be the line through the origin perpendicular to $$L_1$$. We need to find the locus of intersection of $$L_1$$ and $$L_2$$ in the form $$(x^2 + y^2)^2 = \alpha x^2 + \beta y^2$$, and compute $$\alpha + \beta$$.

First, for the hyperbola $$\frac{x^2}{16} - \frac{y^2}{4} = 1$$ (where $$a^2 = 16, b^2 = 4$$), a tangent with slope $$m$$ is given by the line $$y = mx + c$$, subject to the condition $$c^2 = 16m^2 - 4$$.

Next, the line through the origin perpendicular to $$L_1$$ has slope $$-\tfrac{1}{m}$$, so it is described by

$$L_2: y = -\frac{x}{m}$$.

Now let $$(h,k)$$ be the point of intersection of $$L_1$$ and $$L_2$$. From $$L_2$$ we have $$k = -\tfrac{h}{m}$$, or equivalently $$h = -mk$$. Substituting into $$L_1$$ gives

$$k = m(-mk) + c \implies k(1 + m^2) = c \implies k = \frac{c}{1 + m^2},$$

and hence

$$h = -mk = \frac{-mc}{1 + m^2}.$$

Since

$$h^2 + k^2 = \frac{m^2c^2 + c^2}{(1+m^2)^2} = \frac{c^2}{1+m^2},$$

we can express $$m^2$$ and $$c^2$$ in terms of $$h, k$$. From $$h = -mk$$ we obtain $$m = -\frac{h}{k}$$, so $$m^2 = \frac{h^2}{k^2}$$. Substituting into the previous relation yields

$$c^2 = (h^2 + k^2)(1 + m^2) = (h^2 + k^2)\cdot\frac{h^2 + k^2}{k^2} = \frac{(h^2 + k^2)^2}{k^2}.$$

Substituting these into the tangency condition $$c^2 = 16m^2 - 4$$ gives

$$\frac{(h^2 + k^2)^2}{k^2} = 16\cdot\frac{h^2}{k^2} - 4.$$

Multiplying both sides by $$k^2$$ leads to

$$ (h^2 + k^2)^2 = 16h^2 - 4k^2. $$

Comparing with the desired form $$(x^2 + y^2)^2 = \alpha x^2 + \beta y^2$$ shows that $$\alpha = 16$$ and $$\beta = -4$$. Therefore,

$$\alpha + \beta = 16 + (-4) = 12.$$

Hence the answer is $$12$$.

The circle is $$x^2 + y^2 - 2x + 2fy + 1 = 0$$. Its center is $$(1, -f)$$ and radius is $$\sqrt{1 + f^2 - 1} = |f|$$.

Since the center $$(1, -f)$$ lies on both diameters, the equations become:

Diameter 1: $$2p(1) - (-f) = 1 \implies 2p + f = 1$$ ... (i)

Diameter 2: $$2(1) + p(-f) = 4p \implies 2 - pf = 4p$$ ... (ii)

From equation (i), we obtain $$f = 1 - 2p$$.

Substituting this into equation (ii) yields:

$$2 - p(1 - 2p) = 4p$$

$$2 - p + 2p^2 = 4p$$

$$2p^2 - 5p + 2 = 0$$

$$(2p - 1)(p - 2) = 0$$

$$p = \dfrac{1}{2} \text{ or } p = 2$$

When $$p = \dfrac{1}{2}$$, then $$f = 1 - 1 = 0$$, so the radius is 0, which is not valid.

On the other hand, if $$p = 2$$, then $$f = 1 - 4 = -3$$ and the center becomes $$(1, 3)$$.

Next, to find the tangent from the center $$(1, 3)$$ to the hyperbola, note that the hyperbola is $$3x^2 - y^2 = 3$$, or equivalently $$x^2 - \dfrac{y^2}{3} = 1$$, so $$a^2 = 1, b^2 = 3$$.

The general equation of a tangent to the hyperbola with slope $$m$$ is

$$y = mx \pm \sqrt{a^2 m^2 - b^2} = mx \pm \sqrt{m^2 - 3}$$

Since this line must pass through the point $$(1, 3)$$, we substitute to get

$$3 = m \pm \sqrt{m^2 - 3}$$

$$3 - m = \pm \sqrt{m^2 - 3}$$

Squaring both sides leads to

$$ (3 - m)^2 = m^2 - 3 $$

$$ 9 - 6m + m^2 = m^2 - 3 $$

$$ 12 = 6m $$

$$ m = 2 $$

Because $$m = 2 > 0$$ and $$\sqrt{m^2 - 3} = 1$$ satisfies $$3 - 2 = 1$$, this solution is valid.

Therefore, the answer is $$m = 2$$.

We have the standard form of the hyperbola as $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ with its centre at the origin and the transverse axis along the $$x$$-axis. For such a hyperbola the eccentricity $$e$$ is connected to the semi-axes by the well-known relation $$e^{2}=1+\dfrac{b^{2}}{a^{2}}.$$

The question tells us that the eccentricity is $$\dfrac{\sqrt5}{2},$$ so first we square it and substitute in the formula:

$$\left(\dfrac{\sqrt5}{2}\right)^{2}=1+\dfrac{b^{2}}{a^{2}}.$$

Hence

$$\dfrac{5}{4}=1+\dfrac{b^{2}}{a^{2}} \;\Longrightarrow\; \dfrac{b^{2}}{a^{2}}=\dfrac{5}{4}-1=\dfrac{1}{4}.$$

So we obtain the proportional relation between the semi-axes,

$$b^{2}=\dfrac{a^{2}}{4}.$$

Now the given point $$P(-2\sqrt6,\;\sqrt3)$$ lies on the hyperbola. Therefore its coordinates satisfy the equation of the curve. Substituting $$x=-2\sqrt6$$ and $$y=\sqrt3$$ we get

$$\dfrac{(-2\sqrt6)^{2}}{a^{2}}-\dfrac{(\sqrt3)^{2}}{b^{2}}=1.$$

Evaluating the squares,

$$\dfrac{4\cdot6}{a^{2}}-\dfrac{3}{b^{2}}=1 \;\Longrightarrow\; \dfrac{24}{a^{2}}-\dfrac{3}{b^{2}}=1.$$

Using the relation $$b^{2}=\dfrac{a^{2}}{4}$$ we replace $$b^{2}$$ in the above equation:

$$\dfrac{24}{a^{2}}-\dfrac{3}{\dfrac{a^{2}}{4}}=1.$$

The denominator of the second fraction simplifies because dividing by $$\dfrac{a^{2}}{4}$$ is the same as multiplying by $$\dfrac{4}{a^{2}}$$:

$$\dfrac{24}{a^{2}}-\dfrac{3\cdot4}{a^{2}}=1 \;\Longrightarrow\; \dfrac{24}{a^{2}}-\dfrac{12}{a^{2}}=1.$$

Combining the numerators gives

$$\dfrac{12}{a^{2}}=1 \;\Longrightarrow\; a^{2}=12.$$

Substituting back,

$$b^{2}=\dfrac{a^{2}}{4}=\dfrac{12}{4}=3.$$

So the explicit equation of the hyperbola is

$$\dfrac{x^{2}}{12}-\dfrac{y^{2}}{3}=1.$$

Next we need the tangent at the point $$P(x_{1},y_{1})=(-2\sqrt6,\;\sqrt3).$$ For a hyperbola in the above form the point-form of the tangent is stated as

$$\dfrac{xx_{1}}{a^{2}}-\dfrac{yy_{1}}{b^{2}}=1.$$

Substituting $$x_{1}=-2\sqrt6$$, $$y_{1}=\sqrt3$$, $$a^{2}=12$$ and $$b^{2}=3$$ gives

$$\dfrac{x(-2\sqrt6)}{12}-\dfrac{y(\sqrt3)}{3}=1.$$

Simplifying each fraction step by step,

$$-\dfrac{\sqrt6}{6}\,x-\dfrac{\sqrt3}{3}\,y=1.$$

To clear the denominators we multiply every term by $$6$$:

$$-\sqrt6\,x-2\sqrt3\,y=6.$$

Moving every term to one side and multiplying by $$-1$$ for convenience, we may also write the tangent as

$$\sqrt6\,x+2\sqrt3\,y+6=0.$$

The conjugate axis of this hyperbola is the line $$x=0$$ (the $$y$$-axis). To find the point $$Q$$ where the tangent meets this axis, we set $$x=0$$ in the tangent equation:

$$\sqrt6\cdot0+2\sqrt3\,y+6=0 \;\Longrightarrow\; 2\sqrt3\,y=-6 \;\Longrightarrow\; y=-\dfrac{6}{2\sqrt3}=-\dfrac{3}{\sqrt3}=-\sqrt3.$$

Thus

$$Q=(0,\,-\sqrt3).$$

Now we turn to the normal at the same point $$P$$. First we determine the slope of the tangent. Writing the earlier tangent form $$-\sqrt6\,x-2\sqrt3\,y=6$$ in $$y=mx+c$$ form,

$$2\sqrt3\,y=-\sqrt6\,x-6 \;\Longrightarrow\; y=-\dfrac{\sqrt6}{2\sqrt3}\,x-\dfrac{6}{2\sqrt3}.$$

Because $$\dfrac{\sqrt6}{\sqrt3}=\sqrt2$$, we get

$$y=-\dfrac{\sqrt2}{2}\,x-\dfrac{3}{\sqrt3}.$$

Hence the slope of the tangent is

$$m_{\text{tan}}=-\dfrac{\sqrt2}{2}.$$

The slope of the normal is the negative reciprocal, so

$$m_{\text{norm}}=\dfrac{1}{-\dfrac{\sqrt2}{2}}=-\dfrac{2}{\sqrt2}=+\sqrt2.$$ (The double negative disappears, giving a positive value.)

The normal passing through $$P(-2\sqrt6,\;\sqrt3)$$ with slope $$\sqrt2$$ is written via the point-slope formula $$y-y_{1}=m(x-x_{1})$$ as

$$y-\sqrt3=\sqrt2\,(x+2\sqrt6).$$

To find its intersection $$R$$ with the conjugate axis $$x=0$$ we substitute $$x=0$$:

$$y-\sqrt3=\sqrt2\,(0+2\sqrt6)=\sqrt2\cdot2\sqrt6=2\sqrt{12}=2\cdot2\sqrt3=4\sqrt3.$$

Therefore

$$y=\sqrt3+4\sqrt3=5\sqrt3$$

and we obtain

$$R=(0,\;5\sqrt3).$$

Both $$Q$$ and $$R$$ lie on the $$y$$-axis, so their distance is simply the absolute difference of their $$y$$-coordinates:

$$QR=\left|\,5\sqrt3-(-\sqrt3)\right|=|6\sqrt3|=6\sqrt3.$$

Hence, the correct answer is Option D.

For the ellipse $$\frac{x^2}{25} + \frac{y^2}{16} = 1$$, we have $$a = 5$$, $$b = 4$$. The eccentricity is $$e_1 = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$$, and the foci are at $$(\pm 3, 0)$$.

Let the hyperbola be $$\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$$ with eccentricity $$e_2$$. Since the product of eccentricities is 1, we have $$e_1 \cdot e_2 = 1$$, giving $$e_2 = \frac{5}{3}$$.

The hyperbola passes through the foci of the ellipse, so substituting $$(3, 0)$$: $$\frac{9}{A^2} = 1$$, giving $$A^2 = 9$$.

For the hyperbola, $$B^2 = A^2(e_2^2 - 1) = 9\left(\frac{25}{9} - 1\right) = 9 \cdot \frac{16}{9} = 16$$.

Therefore, the equation of the hyperbola is $$\frac{x^2}{9} - \frac{y^2}{16} = 1$$.

Write the hyperbola as $$\frac{x^2}{3}-\frac{y^2}{3}=1$$

So,

$$a^2=3,\qquad b^2=3$$

The line $$2x+y=k$$ becomes $$y=-2x+k$$

Therefore,

$$m=-2,\qquad c=k$$

For the line $$y=mx+c$$ to touch the hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$ the condition is

$$c^2=a^2m^2-b^2$$

Substituting the values,

$$k^2=3(-2)^2-3$$

$$=12-3=9$$

Since

$$k>0,$$ 

$$k=3$$

So the tangent line is $$y=-2x+3$$

Now compare the parabola $$y^2=\alpha x$$ with $$y^2=4ax$$

This gives $$4a=\alpha$$

For the parabola $$y^2=4ax,$$ the tangent with slope $$m$$ is $$y=mx+\frac{a}{m}$$

Comparing with $$y=-2x+3,$$

$$3=\frac{a}{-2}$$

$$a=-6$$

Therefore,

$$\alpha=4a=-24$$

The hyperbola is $$\frac{x^2}{4} - \frac{y^2}{2} = 1$$, so $$a^2 = 4$$, $$b^2 = 2$$, $$c^2 = a^2 + b^2 = 6$$, and $$c = \sqrt{6}$$. The foci are at $$(\pm\sqrt{6}, 0)$$. Since $$P = (4, \sqrt{6})$$ has $$x > 0$$, the nearer focus is $$F = (\sqrt{6}, 0)$$.

The tangent at $$P(4, \sqrt{6})$$ to $$\frac{x^2}{4} - \frac{y^2}{2} = 1$$ is $$\frac{4x}{4} - \frac{\sqrt{6}\,y}{2} = 1$$, i.e., $$x - \frac{\sqrt{6}}{2}y = 1$$. Setting $$y = 0$$ gives the x-intercept $$Q = (1, 0)$$.

The right latus rectum is the vertical line $$x = \sqrt{6}$$. Substituting into the tangent equation: $$\sqrt{6} - \frac{\sqrt{6}}{2}y = 1$$, so $$\frac{\sqrt{6}}{2}y = \sqrt{6} - 1$$, giving $$y = \frac{2(\sqrt{6}-1)}{\sqrt{6}} = 2 - \frac{2}{\sqrt{6}}$$. Thus $$R = \left(\sqrt{6},\; 2 - \frac{2}{\sqrt{6}}\right)$$.

We have $$Q = (1, 0)$$, $$F = (\sqrt{6}, 0)$$, and $$R = \left(\sqrt{6}, 2 - \frac{2}{\sqrt{6}}\right)$$. Since $$Q$$ and $$F$$ both lie on the x-axis, the base $$QF$$ has length $$\sqrt{6} - 1$$. The height is the y-coordinate of $$R$$, which is $$2 - \frac{2}{\sqrt{6}}$$. So the area is $$\frac{1}{2}(\sqrt{6}-1)\left(2 - \frac{2}{\sqrt{6}}\right) = \frac{1}{2}(\sqrt{6}-1) \cdot \frac{2(\sqrt{6}-1)}{\sqrt{6}} = \frac{(\sqrt{6}-1)^2}{\sqrt{6}} = \frac{6 - 2\sqrt{6} + 1}{\sqrt{6}} = \frac{7 - 2\sqrt{6}}{\sqrt{6}} = \frac{7}{\sqrt{6}} - 2$$.

The two lines are $$\sqrt{3}\,kx + ky - 4\sqrt{3} = 0$$ and $$\sqrt{3}\,x - y - 4\sqrt{3}\,k = 0$$.

From the second equation, $$y = \sqrt{3}\,x - 4\sqrt{3}\,k$$, which gives $$k = \frac{\sqrt{3}\,x - y}{4\sqrt{3}}$$.

Substituting into the first equation: $$\sqrt{3}\left(\frac{\sqrt{3}\,x - y}{4\sqrt{3}}\right)x + \left(\frac{\sqrt{3}\,x - y}{4\sqrt{3}}\right)y = 4\sqrt{3}$$.

Simplifying the left side: $$\frac{x(\sqrt{3}\,x - y)}{4} + \frac{y(\sqrt{3}\,x - y)}{4\sqrt{3}} = 4\sqrt{3}$$.

Multiplying throughout by $$4\sqrt{3}$$: $$\sqrt{3}\,x(\sqrt{3}\,x - y) + y(\sqrt{3}\,x - y) = 48$$.

Expanding: $$3x^2 - \sqrt{3}\,xy + \sqrt{3}\,xy - y^2 = 48$$, which simplifies to $$3x^2 - y^2 = 48$$.

Dividing by 48: $$\frac{x^2}{16} - \frac{y^2}{48} = 1$$.

This is a hyperbola with $$a^2 = 16$$ and $$b^2 = 48$$. The eccentricity is $$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{48}{16}} = \sqrt{1 + 3} = \sqrt{4} = 2$$.

Therefore, the eccentricity of the conic is $$2$$.

The hyperbola is $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ and the two points on it are

$$$P\bigl(a\sec\theta,\;b\tan\theta\bigr),\qquad Q\bigl(a\sec\phi,\;b\tan\phi\bigr),\qquad\text{with }\;\theta+\phi=\dfrac{\pi}{2}.$$$

For the hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ the slope of the tangent at a general point $$(x_{1},y_{1})$$ is obtained by implicit differentiation:

$$$\frac{2x_{1}}{a^{2}}-\frac{2y_{1}}{b^{2}}\frac{dy}{dx}=0 \;\Longrightarrow\; \frac{dy}{dx}=\frac{b^{2}x_{1}}{a^{2}y_{1}}.$$$

Hence the slope of the normal is the negative reciprocal:

$$m_{n}=-\frac{a^{2}y_{1}}{b^{2}x_{1}}.$$

Normal at $$P$$

For $$P(a\sec\theta,\;b\tan\theta)$$ we insert $$x_{1}=a\sec\theta,\;y_{1}=b\tan\theta$$ to get

$$$m_{1}=-\frac{a^{2}(b\tan\theta)}{b^{2}(a\sec\theta)} =-\frac{ab\tan\theta}{b^{2}\sec\theta} =-\frac{a}{b}\,\frac{\tan\theta}{\sec\theta} =-\frac{a}{b}\sin\theta.$$$

Thus the normal at $$P$$ is

$$$y-b\tan\theta =-\frac{a}{b}\sin\theta\;\bigl(x-a\sec\theta\bigr). \quad -(1)$$$

Normal at $$Q$$

Because $$\phi=\dfrac{\pi}{2}-\theta,$$ we have $$\sin\phi=\cos\theta$$, $$\; \tan\phi=\cot\theta$$, $$\; \sec\phi=\csc\theta.$$ The slope of the normal at $$Q$$ becomes

$$$m_{2} =-\frac{a^{2}(b\tan\phi)}{b^{2}(a\sec\phi)} =-\frac{a}{b}\frac{\tan\phi}{\sec\phi} =-\frac{a}{b}\sin\phi =-\frac{a}{b}\cos\theta.$$$

Therefore the normal at $$Q$$ is

$$$y-b\tan\phi =-\frac{a}{b}\cos\theta\;\bigl(x-a\sec\phi\bigr). \quad -(2)$$$

Point of intersection of the two normals

Let the intersection be $$(X,Y).$$ Using (1):

$$$Y=b\tan\theta-\frac{a}{b}\sin\theta\,(X-a\sec\theta). \quad -(3)$$$

Using (2) and substituting the trigonometric equivalents:

$$$Y=b\cot\theta-\frac{a}{b}\cos\theta\, \bigl(X-a\csc\theta\bigr). \quad -(4)$$$

Equating the right-hand sides of (3) and (4):

$$$b\tan\theta-\frac{a}{b}\sin\theta\,(X-a\sec\theta) =b\cot\theta-\frac{a}{b}\cos\theta\, \bigl(X-a\csc\theta\bigr).$$$

We first isolate $$X$$. Write the equation as

$$$\left(-\frac{a}{b}\sin\theta+\frac{a}{b}\cos\theta\right)X =b\cot\theta-b\tan\theta +\frac{a^{2}}{b}\sin\theta\sec\theta -\frac{a^{2}}{b}\cos\theta\csc\theta.$$$

Simplifying step by step:

Slope difference (denominator):

$$$m_{1}-m_{2} =-\frac{a}{b}\sin\theta-\Bigl(-\frac{a}{b}\cos\theta\Bigr) =-\frac{a}{b}\bigl(\sin\theta-\cos\theta\bigr).$$$

Numerator:

$$$b\bigl(\cot\theta-\tan\theta\bigr) +\frac{a^{2}}{b}\bigl(\cot\theta-\tan\theta\bigr) =\bigl(\cot\theta-\tan\theta\bigr)\frac{a^{2}+b^{2}}{b}.$$$

Hence

$$$X =\frac{\bigl(\cot\theta-\tan\theta\bigr)\dfrac{a^{2}+b^{2}}{b}} {-\dfrac{a}{b}\bigl(\sin\theta-\cos\theta\bigr)} =-\frac{a^{2}+b^{2}}{a}\; \frac{\cot\theta-\tan\theta}{\sin\theta-\cos\theta}.$$$ Call

$$$R=\frac{\cot\theta-\tan\theta}{\sin\theta-\cos\theta},\qquad \text{so that}\qquad X=-\frac{a^{2}+b^{2}}{a}\,R. \quad -(5)$$$

Value of $$R$$

$$$\cot\theta-\tan\theta =\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta} =\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos\theta} =\frac{\cos2\theta}{\sin\theta\cos\theta}$$$

and

$$\sin\theta-\cos\theta =-(\cos\theta-\sin\theta).$$

Hence

$$$R =\frac{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)} {-(\sin\theta\cos\theta)(\cos\theta-\sin\theta)} =-\frac{\cos\theta+\sin\theta}{\sin\theta\cos\theta}. \quad -(6)$$$

Now the ordinate $$Y$$

From (3):

$$$Y =b\tan\theta-\frac{a}{b}\sin\theta\Bigl(X-a\sec\theta\Bigr).$$$

Using (5):

$$$-\frac{a}{b}\sin\theta\,X =-\frac{a}{b}\sin\theta\Bigl(-\frac{a^{2}+b^{2}}{a}R\Bigr) =\frac{a^{2}+b^{2}}{b}\sin\theta\,R.$$$

Also

$$$-\frac{a}{b}\sin\theta\bigl(-a\sec\theta\bigr) =\frac{a^{2}}{b}\sin\theta\sec\theta =\frac{a^{2}}{b}\tan\theta.$$$

Therefore

$$$Y =b\tan\theta+\frac{a^{2}+b^{2}}{b}\sin\theta\,R+\frac{a^{2}}{b}\tan\theta =\frac{a^{2}+b^{2}}{b}\tan\theta+\frac{a^{2}+b^{2}}{b}\sin\theta\,R.$$$

Factor $$\dfrac{a^{2}+b^{2}}{b}$$:

$$$Y=\frac{a^{2}+b^{2}}{b}\Bigl[\tan\theta+\sin\theta\,R\Bigr]. \quad -(7)$$$

The bracketed constant

Using $$R$$ from (6):

$$$\sin\theta\,R =\sin\theta\Bigl[-\frac{\cos\theta+\sin\theta}{\sin\theta\cos\theta}\Bigr] =-\frac{\cos\theta+\sin\theta}{\cos\theta}.$$$

Hence

$$$\tan\theta+\sin\theta\,R =\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta+\sin\theta}{\cos\theta} =\frac{\sin\theta-\cos\theta-\sin\theta}{\cos\theta} =-\frac{\cos\theta}{\cos\theta} =-1.$$$

Thus the ordinate simplifies to

$$$Y =\frac{a^{2}+b^{2}}{b}\,(-1) =-\frac{a^{2}+b^{2}}{b} =-2\left(\frac{a^{2}+b^{2}}{2b}\right).$$$

Comparing with the given form $$Y=-k\left(\dfrac{a^{2}+b^{2}}{2b}\right)$$ we immediately read

$$k=2.$$

So, the answer is $$2$$.

We are given the hyperbola $$\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$$ whose standard form is $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ with $$a^{2}=4$$ and $$b^{2}=2.$$

For this hyperbola the point-form of the tangent at the point $$(x_{1},y_{1})$$ on the curve is, by the standard formula,

$$\frac{x\,x_{1}}{a^{2}}-\frac{y\,y_{1}}{b^{2}}=1.$$

Substituting $$a^{2}=4$$ and $$b^{2}=2$$ we obtain the required tangent equation:

$$\frac{x\,x_{1}}{4}-\frac{y\,y_{1}}{2}=1.$$

Now we rearrange this equation in the slope-intercept form so that the slope can be compared with the given line. First multiply every term by $$4$$ to clear the denominators:

$$x\,x_{1}-2y\,y_{1}=4.$$

Next isolate the $$y$$-term:

$$-2y\,y_{1}=4-x\,x_{1}.$$

Divide both sides by $$-2y_{1}:$$

$$y=\frac{x\,x_{1}}{2y_{1}}-\frac{4}{2y_{1}}.$$

Simplifying, we get

$$y=\frac{x_{1}}{2y_{1}}\,x-\frac{2}{y_{1}}.$$

Thus the slope $$m$$ of the tangent line is

$$m=\frac{x_{1}}{2y_{1}}.$$

We are told that this tangent is parallel to the straight line $$2x-y=0.$$ Writing that line in slope-intercept form gives $$y=2x,$$ whose slope is $$m=2.$$ Because parallel lines have equal slopes, we equate the two slopes:

$$\frac{x_{1}}{2y_{1}}=2.$$

Multiplying both sides by $$2y_{1}$$ yields

$$x_{1}=4y_{1}.$$

Since $$(x_{1},y_{1})$$ lies on the given hyperbola, it must satisfy its equation. Substituting $$x_{1}=4y_{1}$$ into

$$\frac{x_{1}^{2}}{4}-\frac{y_{1}^{2}}{2}=1$$

gives

$$\frac{(4y_{1})^{2}}{4}-\frac{y_{1}^{2}}{2}=1.$$

Compute the first term:

$$\frac{16y_{1}^{2}}{4}-\frac{y_{1}^{2}}{2}=1,$$

which simplifies to

$$4y_{1}^{2}-\frac{y_{1}^{2}}{2}=1.$$

Write $$4y_{1}^{2}$$ with a denominator of $$2$$ so we can combine like terms:

$$\frac{8y_{1}^{2}}{2}-\frac{y_{1}^{2}}{2}=1.$$

This gives

$$\frac{7y_{1}^{2}}{2}=1.$$

Multiplying both sides by $$\frac{2}{7}$$ produces

$$y_{1}^{2}=\frac{2}{7}.$$

Using $$x_{1}=4y_{1},$$ we have

$$x_{1}^{2}=16y_{1}^{2}=16\left(\frac{2}{7}\right)=\frac{32}{7}.$$

Now we compute the required expression $$x_{1}^{2}+5y_{1}^{2}:$$

$$x_{1}^{2}+5y_{1}^{2}=\frac{32}{7}+5\left(\frac{2}{7}\right)=\frac{32}{7}+\frac{10}{7}=\frac{42}{7}=6.$$

Hence, the correct answer is Option A.

We have been told that the vertices of the hyperbola are at $$(\pm 6,0)$$. For a rectangular Cartesian frame whose origin is at the centre of the hyperbola and whose transverse axis lies along the $$x$$-axis, the standard form of a hyperbola is stated first as

$$\frac{x^{2}}{a^{2}}\;-\;\frac{y^{2}}{b^{2}} \;=\;1,$$

where $$a$$ is the semi-transverse axis. Because the vertices are at $$(\pm a,0)=(\pm 6,0)$$, we immediately identify $$a=6$$, so $$a^{2}=36$$. Thus for our curve the partially known equation is

$$\frac{x^{2}}{36}\;-\;\frac{y^{2}}{b^{2}} \;=\;1.$$

Now the point $$P(10,16)$$ lies on the hyperbola, so it must satisfy the equation. Substituting $$x=10$$ and $$y=16$$ gives

$$\frac{10^{2}}{36}\;-\;\frac{16^{2}}{b^{2}} \;=\;1.$$

Simplifying each term step by step, we get

$$\frac{100}{36}\;-\;\frac{256}{b^{2}} \;=\;1,$$

$$\frac{25}{9}\;-\;\frac{256}{b^{2}} \;=\;1.$$

Moving the second fraction to the right hand side and the constant $$1$$ to the left, we write

$$-\;\frac{256}{b^{2}} \;=\;1-\frac{25}{9},$$

$$-\;\frac{256}{b^{2}} \;=\;\frac{9}{9}-\frac{25}{9} \;=\;-\frac{16}{9}.$$

Removing the minus signs from both sides, we have

$$\frac{256}{b^{2}} \;=\;\frac{16}{9}.$$

Cross-multiplying gives

$$256 \times 9 \;=\;16\,b^{2},$$

$$b^{2} \;=\;\frac{256 \times 9}{16}.$$

Since $$256/16 = 16,$$ it follows that

$$b^{2}=16 \times 9 =144.$$

So the complete equation of the hyperbola is now fixed as

$$\frac{x^{2}}{36}\;-\;\frac{y^{2}}{144} \;=\;1.$$

To find the normal, we must first obtain the tangent at the given point. The general tangent to the hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ at a point $$(x_{1},y_{1})$$ on it is stated by the standard formula

$$\frac{xx_{1}}{a^{2}}\;-\;\frac{yy_{1}}{b^{2}} \;=\;1.$$

Using $$(x_{1},y_{1})=(10,16)$$ and $$a^{2}=36,\;b^{2}=144,$$ we substitute:

$$\frac{x\,(10)}{36}\;-\;\frac{y\,(16)}{144} \;=\;1.$$

Simplifying the numerical coefficients, we notice that

$$\frac{10}{36}=\frac{5}{18},\qquad \frac{16}{144}=\frac{1}{9},$$

so the tangent equation becomes

$$\frac{5x}{18}\;-\;\frac{y}{9} \;=\;1.$$

To obtain a simpler linear form, multiply every term by $$18$$:

$$5x\;-\;2y \;=\;18.$$

Writing this line in the slope-intercept form $$y=mx+c$$, we shift terms:

$$-\,2y = 18-5x,$$

$$2y = 5x-18,$$

$$y = \frac{5}{2}\,x - 9.$$

Thus the slope of the tangent is

$$m_{\text{tangent}} = \frac{5}{2}.$$

For the normal, we use the fact that the product of the slopes of a line and its normal is $$-1$$. Therefore, stating the relation $$m_{\text{tangent}}\;m_{\text{normal}} = -1,$$ we find

$$m_{\text{normal}} = -\,\frac{1}{m_{\text{tangent}}} = -\,\frac{1}{\frac{5}{2}} = -\,\frac{2}{5}.$$

Finally, we write the equation of the normal passing through $$P(10,16)$$ with slope $$-\dfrac{2}{5}$$ using the point-slope form $$y - y_{1} = m(x - x_{1})$$:

$$y - 16 = -\,\frac{2}{5}\,(x - 10).$$

Multiplying by $$5$$ to clear the denominator, we obtain

$$5(y - 16) = -2(x - 10),$$

$$5y - 80 = -2x + 20.$$

Collecting terms on one side, we write

$$2x + 5y - 100 = 0.$$

Or, presenting it more cleanly,

$$2x + 5y = 100.$$

This matches the option offered as $$2x + 5y = 100$$.

Hence, the correct answer is Option B.

We are told that the ellipse is $$3x^2+4y^2=12$$.

First we convert this ellipse to its standard form. Dividing every term by $$12$$ we obtain

$$\frac{x^2}{4}+\frac{y^2}{3}=1.$$

In the standard ellipse $$\dfrac{x^2}{a_e^2}+\dfrac{y^2}{b_e^2}=1$$ we have

$$a_e^2=4,\qquad b_e^2=3.$$

The focal distance for an ellipse is given by the relation

$$c_e^2=a_e^2-b_e^2.$$

Substituting the obtained values,

$$c_e^2=4-3=1\;\;\Longrightarrow\;\;c_e=1.$$

Hence the foci of the ellipse are $$\bigl(\pm1,0\bigr).$$

Because the required hyperbola has the same foci, it must also be centred at the origin with its transverse axis along the $$x$$-axis. Let the hyperbola be

$$\frac{x^2}{a_h^2}-\frac{y^2}{b_h^2}=1.$$

We are further told that the transverse axis of the hyperbola has length $$\sqrt2$$. The transverse axis length equals $$2a_h$$, so we write

$$2a_h=\sqrt2\;\;\Longrightarrow\;\;a_h=\frac{\sqrt2}{2}=\frac1{\sqrt2}.$$

For a hyperbola the focal distance satisfies the well-known formula

$$c_h^2=a_h^2+b_h^2.$$

Here the common foci lie at a distance $$c_h=1$$ from the origin. Substituting the known values,

$$1^2=\left(\frac1{\sqrt2}\right)^2+b_h^2 \;\;\Longrightarrow\;\; 1=\frac12+b_h^2 \;\;\Longrightarrow\;\; b_h^2=1-\frac12=\frac12.$$

Thus the explicit equation of the hyperbola becomes

$$\frac{x^2}{\dfrac12}-\frac{y^2}{\dfrac12}=1.$$

Multiplying numerator and denominator in each fraction by $$2$$ gives

$$2x^2-2y^2=1,$$

or, equivalently,

$$x^2-y^2=\frac12.$$

Now we simply check which of the four given points fails to satisfy this equation.

Option A: $$\left(\dfrac1{\sqrt2},0\right)$$

$$2\!\left(\dfrac1{\sqrt2}\right)^2-2(0)^2 =2\!\left(\frac12\right)-0 =1.$$

The left side equals $$1$$, so the point lies on the hyperbola.

Option B: $$\left(-\sqrt{\dfrac32},1\right)$$

$$2\!\left(\sqrt{\dfrac32}\right)^2-2(1)^2 =2\!\left(\dfrac32\right)-2 =3-2 =1.$$

The equation is satisfied; the point lies on the hyperbola.

Option C: $$\left(1,-\dfrac1{\sqrt2}\right)$$

$$2(1)^2-2\!\left(-\dfrac1{\sqrt2}\right)^2 =2-2\!\left(\dfrac12\right) =2-1 =1.$$

This point also lies on the hyperbola.

Option D: $$\left(\sqrt{\dfrac32},\dfrac1{\sqrt2}\right)$$

$$2\!\left(\sqrt{\dfrac32}\right)^2-2\!\left(\dfrac1{\sqrt2}\right)^2 =2\!\left(\dfrac32\right)-2\!\left(\dfrac12\right) =3-1 =2\neq1.$$

Here the left side is $$2$$, not $$1$$, therefore this point does not lie on the hyperbola.

Hence, the correct answer is Option 4.

First we note that the given line is $$y = mx + c$$. It must touch the circle $$x^2 + y^2 = 36$$ and also touch the hyperbola $$\dfrac{x^2}{100} - \dfrac{y^2}{64} = 1$$. We shall extract one equation for $$c$$ from each tangency condition and then combine them.

For the circle, its centre is clearly $$(0,\,0)$$ and its radius is $$6$$ because $$x^2 + y^2 = 36$$ can be written as $$x^2 + y^2 = 6^2$$. A straight-line $$y = mx + c$$ touches a circle when the perpendicular distance from the centre to the line equals the radius. The distance of $$(0,\,0)$$ from $$y = mx + c$$ is

$$ \frac{|c|}{\sqrt{1 + m^{2}}}. $$

Setting this distance equal to the radius $$6$$, we have

$$ \frac{|c|}{\sqrt{1+m^{2}}}=6 \quad\Longrightarrow\quad c^{2}=36\,(1+m^{2}). \quad -(1) $$

Now we handle the hyperbola. For a rectangular hyperbola $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$, a straight-line with slope $$m$$ is a tangent precisely when it can be written as

$$ y = mx \pm \sqrt{a^{2}m^{2}-b^{2}}. $$

Comparing with $$y = mx + c$$, we identify

$$ c=\pm\sqrt{a^{2}m^{2}-b^{2}} \quad\Longrightarrow\quad c^{2}=a^{2}m^{2}-b^{2}. \quad -(2) $$

For our hyperbola we have $$a^{2}=100$$ and $$b^{2}=64$$, so (2) becomes

$$ c^{2}=100m^{2}-64. \quad -(2') $$

Since the very same $$c$$ must satisfy both (1) and (2′), we equate the right-hand sides:

$$ 36\,(1+m^{2}) = 100m^{2}-64. $$

Expanding the left side and then moving every term to one side, we get

$$ 36 + 36m^{2} = 100m^{2}-64 \\ 36 + 36m^{2} - 100m^{2} + 64 = 0 \\ 100 - 64m^{2} = 0. $$

Re-arranging gives

$$ 64m^{2}=100 \quad\Longrightarrow\quad m^{2}=\frac{100}{64}=\frac{25}{16}. $$

Thus $$m=\pm\dfrac{5}{4}$$. Although the sign of $$m$$ is fixed by the particular tangent chosen, any relation we derive for $$c$$ will involve only $$m^{2}$$, so the sign is immaterial for the present purpose.

We now substitute $$m^{2}=\dfrac{25}{16}$$ into (2′) to obtain $$c^{2}$$:

$$ c^{2}=100\left(\frac{25}{16}\right)-64 =\frac{2500}{16}-\frac{1024}{16} =\frac{1476}{16} =\frac{369}{4}. $$

Multiplying both sides by $$4$$ gives the neat relation

$$ 4c^{2}=369. $$

This coincides exactly with Option C.

Hence, the correct answer is Option C.

The hyperbola is given by $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ and the point $$P(3,3)$$ lies on it. Substituting the coordinates of P, we have

$$\dfrac{3^{2}}{a^{2}}-\dfrac{3^{2}}{b^{2}}=1 \;\;\Longrightarrow\;\; \dfrac{9}{a^{2}}-\dfrac{9}{b^{2}}=1.$$

To find the normal at P, we first need the slope of the tangent. Differentiating the hyperbola implicitly with respect to x, we obtain

$$\dfrac{2x}{a^{2}}-\dfrac{2y}{b^{2}}\dfrac{dy}{dx}=0.$$ Solving for $$\dfrac{dy}{dx}$$ gives

$$\dfrac{dy}{dx}=\dfrac{b^{2}x}{a^{2}y}.$$

At the point $$P(3,3)$$, the slope of the tangent is therefore

$$m_{t}=\left.\dfrac{dy}{dx}\right|_{(3,3)}=\dfrac{b^{2}\cdot3}{a^{2}\cdot3}=\dfrac{b^{2}}{a^{2}}.$$

The slope of the normal is the negative reciprocal of the slope of the tangent, so

$$m_{n}=-\,\dfrac{a^{2}}{b^{2}}.$$

Using the point-slope form, the equation of the normal through $$P(3,3)$$ is

$$y-3=-\dfrac{a^{2}}{b^{2}}\,(x-3).$$

The normal meets the $$x$$-axis where $$y=0$$. Setting $$y=0$$ and solving for $$x$$, we get

$$0-3=-\dfrac{a^{2}}{b^{2}}\,(x-3) \;\;\Longrightarrow\;\; 3=\dfrac{a^{2}}{b^{2}}\,(x-3) \;\;\Longrightarrow\;\; x-3=\dfrac{3b^{2}}{a^{2}} \;\;\Longrightarrow\;\; x=3+\dfrac{3b^{2}}{a^{2}}.$$

But the normal is known to cut the $$x$$-axis at $$(9,0)$$, so

$$9=3+\dfrac{3b^{2}}{a^{2}} \;\;\Longrightarrow\;\; 6=\dfrac{3b^{2}}{a^{2}} \;\;\Longrightarrow\;\; 2=\dfrac{b^{2}}{a^{2}} \;\;\Longrightarrow\;\; b^{2}=2a^{2}.$$

Substituting $$b^{2}=2a^{2}$$ in the earlier relation $$\dfrac{9}{a^{2}}-\dfrac{9}{b^{2}}=1$$, we obtain

$$\dfrac{9}{a^{2}}-\dfrac{9}{2a^{2}}=1 \;\;\Longrightarrow\;\; \dfrac{18-9}{2a^{2}}=1 \;\;\Longrightarrow\;\; \dfrac{9}{2a^{2}}=1 \;\;\Longrightarrow\;\; 2a^{2}=9 \;\;\Longrightarrow\;\; a^{2}=\dfrac{9}{2}.$$

Now, the eccentricity $$e$$ of the hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ satisfies the standard formula $$e^{2}=1+\dfrac{b^{2}}{a^{2}}.$$ Since $$\dfrac{b^{2}}{a^{2}}=2$$, we get

$$e^{2}=1+2=3.$$

Thus the ordered pair is $$\left(a^{2},\,e^{2}\right)=\left(\dfrac{9}{2},\,3\right).$$

Hence, the correct answer is Option A.

For a standard hyperbola with its centre at the origin and transverse axis along the x-axis, we write the equation as $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1.$$ Here

• the length of the conjugate axis is $$2b,$$
• the distance of each focus from the centre is $$c,$$
• the distance between the two foci is therefore $$2c,$$
• the eccentricity is $$e=\dfrac{c}{a},$$
• and the fundamental relation among the semi-axes is $$c^{2}=a^{2}+b^{2}.$$

We are told that the conjugate axis has length 5, so

$$2b = 5 \;\;\Longrightarrow\;\; b = \frac{5}{2}.$$

The distance between the foci is given as 13, hence

$$2c = 13 \;\;\Longrightarrow\;\; c = \frac{13}{2}.$$

By definition of eccentricity we have $$c = ae,$$ so

$$ae = \frac{13}{2} \;\;\Longrightarrow\;\; a = \frac{13}{2e}.$$

Now we invoke the relation $$c^{2}=a^{2}+b^{2}.$$ Substituting $$c=ae$$ and then inserting the known values step by step we get

$$a^{2}e^{2}=a^{2}+b^{2}.$$

Rearranging gives

$$a^{2}(e^{2}-1)=b^{2}.$$

So

$$a^{2}=\frac{b^{2}}{e^{2}-1}.$$

But we already have $$a=\dfrac{13}{2e}$$ from the focus condition. Squaring this expression yields

$$a^{2}=\left(\frac{13}{2e}\right)^{2}=\frac{169}{4e^{2}}.$$

Equating the two expressions for $$a^{2}$$ and substituting $$b^{2}=\left(\dfrac{5}{2}\right)^{2}=\dfrac{25}{4},$$ we obtain

$$\frac{169}{4e^{2}}=\frac{25/4}{\,e^{2}-1\,}.$$

Multiplying both sides by $$4e^{2}(e^{2}-1)$$ eliminates the denominators:

$$169(e^{2}-1)=25e^{2}.$$

Expanding and collecting like terms gives

$$169e^{2}-169=25e^{2},$$

$$169e^{2}-25e^{2}=169,$$

$$144e^{2}=169.$$

Dividing by 144 leads to

$$e^{2}=\frac{169}{144}.$$

Taking the positive square root (because eccentricity is always >1) we get

$$e=\frac{13}{12}.$$

Hence, the correct answer is Option A.

We start by choosing the standard form of a hyperbola whose centre is at the origin and whose transverse axis is along the $$x$$-axis. Hence we write

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1.$$

For this form of hyperbola the two directrices are the vertical lines

$$x=\pm\frac{a}{e},$$

where $$e$$ is the eccentricity. One of the directrices is given in the statement as

$$5x = 4\sqrt{5}\;\;\Longrightarrow\;\;x=\frac{4\sqrt{5}}{5}.$$

Comparing, we must have

$$\frac{a}{e}=\frac{4\sqrt{5}}{5}\quad\Longrightarrow\quad a=\frac{4\sqrt{5}}{5}\,e.$$

Squaring for later convenience gives

$$a^{2}=\left(\frac{4\sqrt{5}}{5}\right)^{2}e^{2} =\frac{16}{5}\,e^{2}.$$

Next, recall the relationship between the semi-axes and the eccentricity of a hyperbola. By definition

$$e=\frac{c}{a},\qquad\text{and}\qquad c^{2}=a^{2}+b^{2}.$$

Eliminating $$c$$ we get

$$e^{2}=1+\frac{b^{2}}{a^{2}} \;\;\Longrightarrow\;\;b^{2}=a^{2}(e^{2}-1).$$

The hyperbola is known to pass through the point $$\bigl(4,\,-2\sqrt{3}\bigr)$$. Substituting $$x=4$$ and $$y=-2\sqrt{3}$$ (note that the sign of $$y$$ is irrelevant because it is squared) in the standard equation gives

$$\frac{4^{2}}{a^{2}}-\frac{(-2\sqrt{3})^{2}}{b^{2}} =1 \;\;\Longrightarrow\;\; \frac{16}{a^{2}}-\frac{12}{b^{2}}=1.$$

Now replace $$b^{2}$$ by $$a^{2}(e^{2}-1)$$:

$$\frac{16}{a^{2}}-\frac{12}{a^{2}(e^{2}-1)}=1.$$

Multiply through by $$a^{2}$$ to clear the denominators:

$$16-\frac{12}{e^{2}-1}=a^{2}.$$

But we already have $$a^{2}=\dfrac{16}{5}e^{2}$$, so equate the two expressions for $$a^{2}$$:

$$\frac{16}{5}e^{2}=16-\frac{12}{e^{2}-1}.$$

Multiply every term by $$5$$ to remove the fraction on the left:

$$16e^{2}=80-\frac{60}{e^{2}-1}.$$

Shift the constant term to the left so that the right side is a single fraction:

$$16e^{2}-80=-\frac{60}{e^{2}-1}.$$

Notice that $$16e^{2}-80=16(e^{2}-5)$$; using that factorisation and multiplying both sides by $$(e^{2}-1)$$ gives

$$16(e^{2}-5)(e^{2}-1)=-60.$$

Expand the product inside the parentheses:

$$(e^{2}-5)(e^{2}-1)=e^{4}-e^{2}-5e^{2}+5=e^{4}-6e^{2}+5.$$

So

$$16\bigl(e^{4}-6e^{2}+5\bigr)=-60.$$

Distribute $$16$$:

$$16e^{4}-96e^{2}+80=-60.$$

Move everything to the left-hand side:

$$16e^{4}-96e^{2}+80+60=0,$$

which simplifies to

$$16e^{4}-96e^{2}+140=0.$$

Finally, divide the whole equation by $$4$$ to make the coefficients smaller:

$$4e^{4}-24e^{2}+35=0.$$

This is the required polynomial equation satisfied by the eccentricity $$e$$. Comparing with the options, we see that this matches Option B.

Hence, the correct answer is Option B.

We want the equation of the tangent to a standard hyperbola that passes through the point $$(4,6)$$ and whose eccentricity is given as $$e = 2.$$

For a standard hyperbola centred at the origin with its transverse axis along the $$x$$-axis, the equation is taken as

$$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1.$$

The eccentricity formula for this hyperbola is first stated:

$$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}.$$

We are told that $$e = 2,$$ so we substitute to obtain

$$2 = \sqrt{1 + \frac{b^{2}}{a^{2}}}.$$

Squaring both sides, we get

$$4 = 1 + \frac{b^{2}}{a^{2}}.$$

Now subtract $$1$$ from each side:

$$\frac{b^{2}}{a^{2}} = 3.$$

So we have the direct relation

$$b^{2} = 3a^{2}.$$

Next, because the given point $$(4,6)$$ lies on the hyperbola, it satisfies its equation. Hence we substitute $$x = 4$$ and $$y = 6$$ into

$$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1.$$

This gives

$$\frac{4^{2}}{a^{2}} - \frac{6^{2}}{b^{2}} = 1.$$

Simplifying the numerators, we write

$$\frac{16}{a^{2}} - \frac{36}{b^{2}} = 1.$$

We already know $$b^{2} = 3a^{2},$$ so we substitute it here:

$$\frac{16}{a^{2}} - \frac{36}{3a^{2}} = 1.$$

The second fraction simplifies because $$\frac{36}{3} = 12,$$ so we have

$$\frac{16}{a^{2}} - \frac{12}{a^{2}} = 1.$$

The numerators on the left combine to give

$$\frac{4}{a^{2}} = 1.$$

Now multiply both sides by $$a^{2}:$$

$$4 = a^{2}.$$

Hence we obtain

$$a^{2} = 4 \quad\text{and}\quad b^{2} = 3a^{2} = 12.$$

The complete equation of the hyperbola is therefore

$$\frac{x^{2}}{4} - \frac{y^{2}}{12} = 1.$$

We are required to write the tangent at the point $$(x_1,y_1) = (4,6).$$ For a hyperbola of the form $$\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1,$$ the equation of the tangent at $$(x_{1},y_{1})$$ is given by the standard formula

$$\frac{xx_{1}}{a^{2}} - \frac{yy_{1}}{b^{2}} = 1.$$

Substituting $$x_{1}=4,\;y_{1}=6,\;a^{2}=4,\;b^{2}=12,$$ we have

$$\frac{x\cdot 4}{4} - \frac{y\cdot 6}{12} = 1.$$

Now each fraction simplifies:

$$x - \frac{y}{2} = 1.$$

To clear the denominator, multiply every term by $$2:$$

$$2x - y = 2.$$

Finally, we rewrite it in the form with all terms on the left-hand side:

$$2x - y - 2 = 0.$$

This exactly matches Option D.

Hence, the correct answer is Option D.

We are given the hyperbola $$\dfrac{x^{2}}{24}-\dfrac{y^{2}}{18}=1$$ and the straight line $$y=mx+7\sqrt{3}$$. For the line to be normal to the hyperbola, two conditions must hold:

1. The slope of the line must equal the slope of the normal to the hyperbola at their point of contact.
2. The point of contact must satisfy both the equation of the hyperbola and that of the line.

First, we differentiate the hyperbola to find the slope of the tangent. Starting with $$\frac{x^{2}}{24}-\frac{y^{2}}{18}=1,$$ we differentiate implicitly with respect to $$x$$:

$$\frac{2x}{24}-\frac{2y}{18}\,\frac{dy}{dx}=0.$$

Simplifying the coefficients, we get

$$\frac{x}{12}-\frac{y}{9}\,\frac{dy}{dx}=0.$$

Solving for $$\dfrac{dy}{dx}$$ (the slope of the tangent),

$$-\frac{y}{9}\,\frac{dy}{dx}=-\frac{x}{12}\quad\Longrightarrow\quad \frac{dy}{dx}=\frac{x}{12}\cdot\frac{9}{y}=\frac{3x}{4y}.$$

The slope of the normal is the negative reciprocal of the slope of the tangent, so

$$m_{\text{normal}}=-\frac{1}{\dfrac{dy}{dx}}=-\frac{4y}{3x}.$$

If the normal passes through the point $$(x_{1},y_{1})$$ on the hyperbola, then the given line $$y=mx+7\sqrt{3}$$ must satisfy

$$m=-\frac{4y_{1}}{3x_{1}}.\qquad(1)$$

Because the point lies on the line as well, we must also have

$$y_{1}=mx_{1}+7\sqrt{3}.\qquad(2)$$

We now solve equations (1) and (2) together. From (1) we express $$y_{1}$$ in terms of $$x_{1}$$:

$$y_{1}=-\frac{3m}{4}\,x_{1}.\qquad(3)$$

Substituting (3) into (2):

$$-\frac{3m}{4}\,x_{1}-m x_{1}=7\sqrt{3}.$$

Combining the coefficients of $$x_{1}$$:

$$\left(-\frac{3}{4}-1\right)m x_{1}=7\sqrt{3}\;\;\Longrightarrow\;\; -\frac{7}{4}m x_{1}=7\sqrt{3}.$$

Dividing both sides by $$-\dfrac{7}{4}m$$, we find

$$x_{1}=\frac{7\sqrt{3}}{-\dfrac{7}{4}m}=-\frac{4\sqrt{3}}{m}.$$

Using (3) for $$y_{1}$$:

$$y_{1}=-\frac{3m}{4}\left(-\frac{4\sqrt{3}}{m}\right)=3\sqrt{3}.$$

Thus the point of contact is $$(x_{1},y_{1})=\left(-\dfrac{4\sqrt{3}}{m},\,3\sqrt{3}\right).$$ This point must satisfy the hyperbola’s equation:

$$\frac{x_{1}^{2}}{24}-\frac{y_{1}^{2}}{18}=1.$$

Computing each term:

$$x_{1}^{2}=\left(-\frac{4\sqrt{3}}{m}\right)^{2}=\frac{48}{m^{2}},\qquad y_{1}^{2}=(3\sqrt{3})^{2}=27.$$

Substituting these values,

$$\frac{\dfrac{48}{m^{2}}}{24}-\frac{27}{18}=1 \;\;\Longrightarrow\;\; \frac{2}{m^{2}}-\frac{3}{2}=1.$$

Taking all terms to one side,

$$\frac{2}{m^{2}}=1+\frac{3}{2}=\frac{5}{2}.$$

Now invert the fraction and solve for $$m^{2}$$:

$$m^{2}=\frac{2}{5}\cdot2=\frac{4}{5}.$$

Hence $$m=\pm\frac{2}{\sqrt{5}}.$$

Among the given choices only the positive value $$\dfrac{2}{\sqrt{5}}$$ appears. Hence, the correct answer is Option D.

We first translate the two curves into their standard forms. For the parabola we have $$y^{2}=12x,$$ which can be written as $$y^{2}=4ax$$ with $$4a=12$$ so $$a=3.$$ Thus the focus of the parabola is at $$(3,0).$$

The hyperbola is $$8x^{2}-y^{2}=8.$$ Dividing by 8 gives $$\frac{x^{2}}{1}-\frac{y^{2}}{8}=1,$$ so it is of the standard form $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ with $$a^{2}=1,\quad b^{2}=8.$$ For a hyperbola the focal distance is obtained from $$c^{2}=a^{2}+b^{2},$$ hence $$c^{2}=1+8=9 \;\Longrightarrow\; c=3.$$ Therefore the two foci of the hyperbola are $$(3,0) \text{ and } (-3,0).$$ We are told that $$S=(3,0)$$ (the focus on the positive $$x$$-axis) and $$S' =(-3,0).$$

We now find the common tangents to both curves.

(i) Tangent to the parabola. For a parabola $$y^{2}=4ax,$$ the slope form of a tangent is $$y=mx+\frac{a}{m}.$$ Substituting $$a=3$$ gives the parabola’s tangent as $$y=mx+\frac{3}{m}\,. \quad -(1)$$

(ii) Tangent to the hyperbola. For a hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1,$$ a straight line $$y=mx+c$$ is tangent when $$c^{2}=a^{2}m^{2}-b^{2}.$$ Here $$a^{2}=1,\; b^{2}=8,$$ so the tangency condition is $$c^{2}=1\cdot m^{2}-8 = m^{2}-8. \quad -(2)$$

Because a common tangent must satisfy both (1) and (2), we must have the same intercept $$c$$ in each. From (1) we see that $$c=\frac{3}{m},$$ and substituting this into (2) gives $$\left(\frac{3}{m}\right)^{2}=m^{2}-8.$$ Carrying out the algebra step by step:

$$\frac{9}{m^{2}} = m^{2}-8,$$ $$9 = m^{4}-8m^{2},$$ $$m^{4}-8m^{2}-9=0.$$

Set $$t=m^{2}$$ to turn it into a quadratic: $$t^{2}-8t-9=0.$$ Using the quadratic formula, $$t=\frac{8\pm\sqrt{64+36}}{2}=\frac{8\pm\sqrt{100}}{2}=\frac{8\pm10}{2}.$$ Thus $$t=9 \quad\text{or}\quad t=-1.$$ Since $$t=m^{2}\ge 0,$$ we accept $$m^{2}=9,$$ giving the two real slopes $$m=3 \quad\text{or}\quad m=-3.$$

For each slope the corresponding $$c$$ is $$c=\frac{3}{m}.$$ Hence for $$m=3,\; c=1,$$ yielding the line $$y=3x+1;$$ for $$m=-3,\; c=-1,$$ yielding the line $$y=-3x-1.$$

These two lines are the common tangents to both curves. Their point of intersection $$P$$ is obtained by solving

$$\begin{cases} y=3x+1,\\[4pt] y=-3x-1. \end{cases}$$

Equating the two expressions for $$y$$ gives $$3x+1 = -3x-1 \;\Longrightarrow\; 6x = -2 \;\Longrightarrow\; x = -\dfrac{1}{3}.$$ Substituting back, $$y = 3\!\left(-\dfrac{1}{3}\right)+1 = -1+1 = 0.$$ Therefore $$P\left(-\dfrac{1}{3},\,0\right).$$

Because $$S(3,0),\;P(-\tfrac{1}{3},0)$$ and $$S'(-3,0)$$ all lie on the $$x$$-axis, we can compute the distances simply from their $$x$$-coordinates.

Distance $$SP = 3 -\!\left(-\dfrac{1}{3}\right) = 3+\dfrac{1}{3} = \dfrac{10}{3},$$ distance $$PS' = -\dfrac{1}{3} -(-3) = -\dfrac{1}{3}+3 = \dfrac{8}{3}.$$

Hence the ratio in which $$P$$ divides $$\overline{SS'}$$ is $$SP : PS' = \dfrac{10}{3} : \dfrac{8}{3} = 10 : 8 = 5 : 4.$$

Hence, the correct answer is Option A.

We are given the hyperbola $$16x^2-9y^2=144$$. To bring it into the standard form, we divide every term by $$144$$:

$$\frac{16x^2}{144}-\frac{9y^2}{144}=1 \;\Longrightarrow\; \frac{x^2}{9}-\frac{y^2}{16}=1.$$

Now the equation is of the form $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ with

$$a^2=9,\; b^2=16 \;\Longrightarrow\; a=3,\; b=4.$$

For a rectangular hyperbola oriented along the $$x$$-axis, the distance of each focus from the centre satisfies the relation

$$c^2=a^2+b^2.$$

Substituting $$a^2=9$$ and $$b^2=16$$ we obtain

$$c^2=9+16=25 \;\Longrightarrow\; c=5.$$

The eccentricity $$e$$ is defined by the formula

$$e=\frac{c}{a}.$$

Hence

$$e=\frac{5}{3}.$$

For the hyperbola $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1,$$ the directrices are given by the pair of vertical lines

$$x=\pm\frac{a}{e}.$$

Substituting $$a=3$$ and $$e=\dfrac{5}{3},$$ we get

$$\frac{a}{e}=\frac{3}{\frac{5}{3}}=\frac{9}{5}.$$

So the two directrices are

$$x=\frac{9}{5}\quad\text{and}\quad x=-\frac{9}{5}.$$

The problem states that the directrix is $$5x+9=0,$$ which can be rewritten as

$$x=-\frac{9}{5}.$$

This matches the negative directrix $$x=-\dfrac{9}{5}.$$ The corresponding focus lies on the same (negative) side of the centre along the transverse axis. The foci for this hyperbola are $$(\pm c,0)=(\pm5,0).$$ Therefore, the required focus is

$$(-5,0).$$

Hence, the correct answer is Option A.

We are given the hyperbola $$\frac{x^{2}}{\cos^{2}\theta}-\frac{y^{2}}{\sin^{2}\theta}=1$$ with the condition $$0<\theta<\frac{\pi}{2}$$ and the extra information that its eccentricity is greater than 2. We must find the possible values of the length of its latus‐rectum.

For any hyperbola in the standard form $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ we know two important formulas:

1. Eccentricity: $$e^{2}=1+\frac{b^{2}}{a^{2}}.$$
2. Length of the latus rectum: $$\text{latus rectum}= \frac{2b^{2}}{a}.$$

Comparing the given equation with the standard form, we can immediately read off

$$a^{2}=\cos^{2}\theta \quad\Longrightarrow\quad a=\cos\theta,$$

$$b^{2}=\sin^{2}\theta.$$

Now we apply the eccentricity formula. Substituting $$a^{2}=\cos^{2}\theta$$ and $$b^{2}=\sin^{2}\theta$$ into $$e^{2}=1+\dfrac{b^{2}}{a^{2}},$$ we obtain

$$e^{2}=1+\frac{\sin^{2}\theta}{\cos^{2}\theta}=1+\tan^{2}\theta.$$

By the Pythagorean identity, $$1+\tan^{2}\theta=\sec^{2}\theta,$$ so

$$e=\sec\theta.$$

The question states that the eccentricity is greater than 2, therefore

$$\sec\theta > 2\quad\Longrightarrow\quad \cos\theta <\frac12.$$

Because $$0<\theta<\frac{\pi}{2},$$ the cosine function decreases from 1 to 0 in this interval, so the inequality $$\cos\theta <\frac12$$ is equivalent to

$$\theta >\cos^{-1}\!\left(\frac12\right)=\frac{\pi}{3}.$$

Hence

$$\frac{\pi}{3}<\theta<\frac{\pi}{2}\quad\Longrightarrow\quad 0<\cos\theta<\frac12.$$

To determine the length of the latus rectum, we use the second formula stated above. Substituting $$b^{2}=\sin^{2}\theta$$ and $$a=\cos\theta,$$ we find

$$\text{latus rectum}= \frac{2\sin^{2}\theta}{\cos\theta}.$$

It is convenient to write everything in terms of $$x=\cos\theta,$$ because $$x$$ varies over the interval $$(0,\tfrac12).$$ Noting that $$\sin^{2}\theta=1-\cos^{2}\theta=1-x^{2},$$ we obtain

$$\text{latus rectum}=2\,\frac{1-x^{2}}{x}=2\!\left(\frac1x-x\right).$$

Define the function $$f(x)=2\!\left(\frac1x-x\right)$$ for $$x\in(0,\tfrac12).$$ We analyse how $$f(x)$$ behaves on this interval in order to find the possible range of the latus rectum.

The derivative is

$$f'(x)=2\!\left(-\frac1{x^{2}}-1\right)=-\frac{2}{x^{2}}-2,$$

which is strictly negative for all positive $$x.$$ Therefore $$f(x)$$ is strictly decreasing as $$x$$ increases. We now look at the two end‐points of the interval:

1. As $$x\to0^{+},$$ we have $$\frac1x\to\infty,$$ so $$f(x)\to\infty.$$

2. At the right‐hand end $$x=\frac12,$$ we get

$$f\!\left(\frac12\right)=2\!\left(\frac1{\frac12}-\frac12\right)=2\,(2-0.5)=2\,(1.5)=3.$$

Because $$f(x)$$ is decreasing, the values of the latus rectum fill exactly the open interval $$(3,\infty).$$

Hence, the correct answer is Option A.

We are asked to write the equation of a tangent to the hyperbola $$4x^2-5y^2=20$$ that is parallel to the straight line $$x-y=2$$.

First we note that a line written as $$x-y=2$$ can be rearranged as $$y=x-2.$$ From this form we immediately read its slope; the coefficient of $$x$$ is $$1$$, so the slope of every line parallel to it is also $$m=1$$.

Now we place the hyperbola in its standard form. Dividing every term of $$4x^2-5y^2=20$$ by $$20$$, we get $$\frac{x^2}{5}-\frac{y^2}{4}=1.$$ Comparing with the general standard form $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$ we identify $$a^2=5 \quad\text{and}\quad b^2=4.$$ Hence $$a=\sqrt5,\qquad b=2.$

The slope form of the tangent to a hyperbola $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$ is a known result; it is

$$y = mx \;\pm\; \sqrt{a^2m^2-b^2}.$$

We explicitly cite this formula so that each subsequent substitution is perfectly clear.

We already found the desired slope $$m=1$$. Substituting $$m=1$$, $$a^2=5$$, and $$b^2=4$$ into the square-root term, we proceed one algebraic step at a time:

$$a^2m^2-b^2 =5\,(1)^2-4 =5-4 =1.$$

The square root of $$1$$ is simply $$1$$, so the tangent lines become

$$y = 1\cdot x \;\pm\; 1,$$

or written more transparently,

$$y = x + 1 \quad\text{and}\quad y = x - 1.$$

To match the style of the answer choices, we transfer each line to the form $$Ax + By + C = 0$$ by moving every term to one side:

For $$y = x + 1$$ we subtract $$y$$ and add $$1$$ to the left:

$$x - y + 1 = 0.$$

For $$y = x - 1$$ we again subtract $$y$$ and now subtract $$1$$ on the left:

$$x - y - 1 = 0.$$

Among the four answer choices given in the problem statement we find only one of these two candidates, namely

$$x - y + 1 = 0.$$

Therefore the required tangent, parallel to the line $$x-y=2$$, is $$x - y + 1 = 0$$.

Hence, the correct answer is Option C.

We are told that the centre of the hyperbola is the origin, its transverse axis lies along the $$x$$-axis and the length of this transverse axis is 4.

For a hyperbola whose transverse axis is along the $$x$$-axis with centre at the origin, the standard equation is

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1.$$

Here $$2a$$ represents the length of the transverse axis. Since the given length is 4, we have

$$2a = 4.$$

Dividing both sides by 2, we get

$$a = 2.$$

So the equation of the required hyperbola can now be written as

$$\frac{x^{2}}{2^{2}}-\frac{y^{2}}{b^{2}}=1,$$

which simplifies to

$$\frac{x^{2}}{4}-\frac{y^{2}}{b^{2}}=1.$$

The hyperbola passes through the point $$(4,\,2)$$. That means when $$x = 4$$ and $$y = 2$$, the left side of the equation must equal 1. Substituting these values, we have

$$\frac{(4)^{2}}{4}-\frac{(2)^{2}}{b^{2}} = 1.$$

Calculating the squares gives

$$\frac{16}{4}-\frac{4}{b^{2}} = 1.$$

Simplifying the first fraction, we obtain

$$4-\frac{4}{b^{2}} = 1.$$

Now we isolate the term containing $$b^{2}$$ by subtracting 1 from both sides:

$$4-\frac{4}{b^{2}}-1 = 0 \quad\Longrightarrow\quad 3-\frac{4}{b^{2}} = 0.$$

Next, we add $$\dfrac{4}{b^{2}}$$ to both sides to remove the negative sign:

$$3 = \frac{4}{b^{2}}.$$

To solve for $$b^{2}$$, we take the reciprocal of both sides:

$$\frac{1}{3} = \frac{b^{2}}{4}.$$

Multiplying both sides by 4, we get

$$b^{2} = \frac{4}{3}.$$

We are now ready to compute the eccentricity $$e$$. For a hyperbola in the form

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1,$$

the formula for the eccentricity is

$$e = \sqrt{1+\frac{b^{2}}{a^{2}}}.$$

We already have $$a^{2}=4$$ and $$b^{2}=\dfrac{4}{3}$$. Substituting these values into the formula gives

$$e = \sqrt{1+\frac{\dfrac{4}{3}}{4}}.$$

Inside the square root, the fraction simplifies as follows:

$$\frac{\dfrac{4}{3}}{4} = \frac{4}{3}\times\frac{1}{4} = \frac{1}{3}.$$

Hence the expression under the radical becomes

$$1 + \frac{1}{3} = \frac{4}{3}.$$

Therefore, we obtain

$$e = \sqrt{\frac{4}{3}}.$$

Taking the square root of the numerator and the denominator separately, we arrive at

$$e = \frac{2}{\sqrt{3}}.$$

Thus the eccentricity of the hyperbola is $$\dfrac{2}{\sqrt{3}}$$.

Hence, the correct answer is Option C.

We are told that the vertices of the required hyperbola are at $$(-2,0)$$ and $$(2,0)$$. For any hyperbola with its transverse (real) axis along the $$x$$-axis, the standard form is

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,$$

where $$\left(\pm a,0\right)$$ are the vertices. Comparing this fact with the given vertices, we at once see that

$$a=2,\qquad a^{2}=4.$$

Next, one focus of the hyperbola is stated to be at $$(-3,0)$$. In the same standard orientation, the foci are at $$\left(\pm c,0\right)$$, so we identify

$$c=3,\qquad c^{2}=9.$$

A basic hyperbola identity connects the semi-transverse axis $$a$$, the semi-conjugate axis $$b$$ and the focal distance $$c$$:

$$c^{2}=a^{2}+b^{2}\qquad\text{(for a hyperbola).}$$

Substituting the already found values, we get

$$9 = 4 + b^{2},$$

so

$$b^{2}=9-4=5.$$

Hence, the complete Cartesian equation of the hyperbola is

$$\frac{x^{2}}{4}-\frac{y^{2}}{5}=1.$$

Now we shall test each of the four given points one by one by substituting their coordinates into this equation. A point lies on the curve only when the left-hand side equals $$1$$.

Option A: $$(6,\,5\sqrt{2})$$

We have

$$\frac{x^{2}}{4}=\frac{6^{2}}{4}=\frac{36}{4}=9,$$

and

$$\frac{y^{2}}{5}=\frac{(5\sqrt{2})^{2}}{5}=\frac{25\cdot 2}{5}=\frac{50}{5}=10.$$

So the left-hand side becomes

$$9-10=-1\neq 1.$$

Therefore $$(6,\,5\sqrt{2})$$ does not satisfy the equation and is not on the hyperbola.

Option B: $$(-6,\,2\sqrt{10})$$

Compute

$$\frac{x^{2}}{4}=\frac{(-6)^{2}}{4}=\frac{36}{4}=9,\qquad \frac{y^{2}}{5}=\frac{(2\sqrt{10})^{2}}{5}=\frac{4\cdot 10}{5}=8.$$

The result is

$$9-8=1,$$

so this point indeed lies on the hyperbola.

Option C: $$(2\sqrt{6},\,5)$$

Compute

$$\frac{x^{2}}{4}=\frac{(2\sqrt{6})^{2}}{4}=\frac{4\cdot 6}{4}=6,\qquad \frac{y^{2}}{5}=\frac{5^{2}}{5}=\frac{25}{5}=5.$$

Thus

$$6-5=1,$$

so this point also lies on the curve.

Option D: $$(4,\,\sqrt{15})$$

Compute

$$\frac{x^{2}}{4}=\frac{4^{2}}{4}=\frac{16}{4}=4,\qquad \frac{y^{2}}{5}=\frac{(\sqrt{15})^{2}}{5}=\frac{15}{5}=3.$$

Hence

$$4-3=1,$$

so this point, too, satisfies the hyperbola’s equation.

Among the four points, only Option A fails to satisfy $$\dfrac{x^{2}}{4}-\dfrac{y^{2}}{5}=1$$ and hence does not belong to the hyperbola.

Hence, the correct answer is Option A.

We have two concurrent straight lines depending on a parameter $$k\neq 0$$:

$$\sqrt{2}\,x-y+4\sqrt{2}\,k=0\qquad\text{and}\qquad \sqrt{2}\,k\,x+k\,y-4\sqrt{2}=0.$$

The task is to find the locus of their point of intersection. Let the required point be $$(x,y)$$. Because the point lies on the first line, we may express $$y$$ in terms of $$x$$ and $$k$$:

$$\sqrt{2}\,x-y+4\sqrt{2}\,k=0\;\Longrightarrow\;y=\sqrt{2}\,x+4\sqrt{2}\,k.$$

Now we substitute this value of $$y$$ in the second line:

$$\sqrt{2}\,k\,x+k\bigl(\sqrt{2}\,x+4\sqrt{2}\,k\bigr)-4\sqrt{2}=0.$$

Simplifying term by term,

$$\sqrt{2}\,k\,x+k\sqrt{2}\,x+4\sqrt{2}\,k^{2}-4\sqrt{2}=0,$$

and the first two terms are identical, so they combine to give

$$2\sqrt{2}\,k\,x+4\sqrt{2}\,k^{2}-4\sqrt{2}=0.$$

Dividing the whole equation by $$2\sqrt{2}$$ (non-zero) yields

$$k\,x+2k^{2}-2=0.$$

This is a quadratic in $$k$$:

$$2k^{2}+xk-2=0.$$

For a quadratic $$ak^{2}+bk+c=0$$, the quadratic formula gives $$k=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$ Applying it with $$a=2,\; b=x,\; c=-2$$ we obtain

$$k=\dfrac{-x\pm\sqrt{x^{2}+16}}{4}.$$

Next we substitute this value of $$k$$ back in the earlier expression for $$y$$:

$$y=\sqrt{2}\,x+4\sqrt{2}\,k =\sqrt{2}\,x+4\sqrt{2}\left(\dfrac{-x\pm\sqrt{x^{2}+16}}{4}\right) =\sqrt{2}\,x+\sqrt{2}\bigl(-x\pm\sqrt{x^{2}+16}\bigr).$$

The $$\sqrt{2}\,x$$ terms cancel, leaving

$$y=\pm\sqrt{2}\,\sqrt{x^{2}+16}.$$

Eliminating the ambiguous sign by squaring, we get

$$y^{2}=2\bigl(x^{2}+16\bigr)=2x^{2}+32.$$

Rearranging the terms,

$$y^{2}-2x^{2}=32.$$

To recognise the conic, we divide by $$32$$:

$$\frac{y^{2}}{32}-\frac{x^{2}}{16}=1.$$

This is clearly of the standard form $$\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1,$$ which represents a hyperbola whose transverse axis is along the $$y$$-axis. Here

$$a^{2}=32\;\Longrightarrow\;a=\sqrt{32}=4\sqrt{2}.$$

The length of the transverse axis of a hyperbola is $$2a$$, so in the present case

$$2a=2\bigl(4\sqrt{2}\bigr)=8\sqrt{2}.$$

Among the given options, this matches the statement “a hyperbola with length of its transverse axis $$8\sqrt{2}$$.”

Hence, the correct answer is Option C.

We are given the hyperbola $$4x^{2}-y^{2}=36$$. Dividing each term by $$36$$ we write it in standard form

$$\frac{x^{2}}{9}-\frac{y^{2}}{36}=1.$$

Hence $$a^{2}=9$$ and $$b^{2}=36$$ for the hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1.$$

For this hyperbola the slope-form (also called the $$m$$-form) of a tangent is

$$y = mx \;\pm\; \sqrt{a^{2}m^{2}-b^{2}}.$$

We know that the two tangents meet at the fixed point $$T(0,3)$$. If the slope of a tangent is $$m$$, its equation through the point $$T(0,3)$$ is obtained from the two-point form:

$$y-3 = m(x-0) \;\Longrightarrow\; y = mx + 3.$$

This very same line must also fit the slope-form, so we equate the constant terms. Comparing

$$y = mx + 3 \quad\text{with}\quad y = mx \;\pm\; \sqrt{a^{2}m^{2}-b^{2}},$$

we require

$$\sqrt{a^{2}m^{2}-b^{2}} = 3 \quad\text{or}\quad \sqrt{a^{2}m^{2}-b^{2}} = -3.$$

Because a square-root is non-negative we take the positive sign and square both sides:

$$a^{2}m^{2}-b^{2} = 9.$$

Substituting $$a^{2}=9$$ and $$b^{2}=36$$ we obtain

$$9m^{2}-36 = 9,$$

$$9m^{2} = 45,$$

$$m^{2} = 5,$$

so

$$m = \sqrt{5}\quad\text{or}\quad m = -\sqrt{5}.$$

Thus the two tangents are

$$\text{(i)}\; y = \sqrt{5}\,x + 3,$$

$$\text{(ii)}\; y = -\sqrt{5}\,x + 3.$$

To find the points of contact $$P$$ and $$Q$$ we intersect each tangent with the hyperbola, and because tangency means a double root, we can proceed directly.

Tangent with slope $$m=\sqrt{5}$$

Substitute $$y = \sqrt{5}\,x + 3$$ into $$4x^{2}-y^{2}=36$$:

$$4x^{2} - \left(\sqrt{5}\,x + 3\right)^{2} = 36.$$

Expand the square:

$$4x^{2} - \left(5x^{2} + 6\sqrt{5}\,x + 9\right) = 36.$$

Simplify term by term:

$$4x^{2} - 5x^{2} - 6\sqrt{5}\,x - 9 = 36,$$

$$-x^{2} - 6\sqrt{5}\,x - 9 = 36,$$

$$-x^{2} - 6\sqrt{5}\,x - 45 = 0.$$

Multiply through by $$-1$$:

$$x^{2} + 6\sqrt{5}\,x + 45 = 0.$$

The discriminant is

$$\Delta = (6\sqrt{5})^{2} - 4\cdot 1 \cdot 45 = 180 - 180 = 0,$$

confirming tangency. The repeated root gives the $$x$$-coordinate of the point of contact:

$$x = -\frac{6\sqrt{5}}{2} = -3\sqrt{5}.$$

Put this value into the tangent equation to get $$y$$:

$$y = \sqrt{5}\,(-3\sqrt{5}) + 3 = -15 + 3 = -12.$$

So

$$P\bigl(-3\sqrt{5},\,-12\bigr).$$

Tangent with slope $$m=-\sqrt{5}$$

Substitute $$y = -\sqrt{5}\,x + 3$$ into $$4x^{2}-y^{2}=36$$:

$$4x^{2} - \left(-\sqrt{5}\,x + 3\right)^{2} = 36.$$

Expand the square:

$$4x^{2} - \left(5x^{2} - 6\sqrt{5}\,x + 9\right) = 36.$$

Simplify term by term:

$$4x^{2} - 5x^{2} + 6\sqrt{5}\,x - 9 = 36,$$

$$-x^{2} + 6\sqrt{5}\,x - 9 = 36,$$

$$-x^{2} + 6\sqrt{5}\,x - 45 = 0.$$

Multiply through by $$-1$$:

$$x^{2} - 6\sqrt{5}\,x + 45 = 0.$$

The discriminant is again zero:

$$\Delta = (-6\sqrt{5})^{2} - 4\cdot 1 \cdot 45 = 180-180=0,$$

so the double root is

$$x = \frac{6\sqrt{5}}{2} = 3\sqrt{5}.$$

The $$y$$-coordinate is

$$y = -\sqrt{5}\,(3\sqrt{5}) + 3 = -15 + 3 = -12.$$

Therefore

$$Q\bigl(3\sqrt{5},\,-12\bigr).$$

We now possess the three vertices of the triangle:

$$P(-3\sqrt{5},-12), \quad Q(3\sqrt{5},-12), \quad T(0,3).$$

The line $$PQ$$ is horizontal because both points share the same $$y$$-coordinate $$-12$$, so

$$\text{base } PQ = |\,3\sqrt{5} - (-3\sqrt{5})\,| = 6\sqrt{5}.$$

The altitude from $$T$$ to the base is the vertical distance between $$y=3$$ and $$y=-12$$:

$$\text{height } = 3 - (-12) = 15.$$

Using the formula for the area of a triangle, $$\text{Area} = \dfrac{1}{2}\times\text{base}\times\text{height},$$ we get

$$\text{Area} = \frac{1}{2}\times (6\sqrt{5}) \times 15 = 3\sqrt{5}\times 15 = 45\sqrt{5}.$$

Hence, the correct answer is Option B.

We are given two straight lines, each involving a real parameter $$t$$:

$$t\,x - 2y - 3t = 0 \qquad\text{and}\qquad x - 2t\,y + 3 = 0.$$

The required locus is the set of all intersection points $$(x,y)$$ obtained when $$t$$ takes every real value. To find this locus we must eliminate $$t$$ from the two equations.

First, from the equation $$t\,x - 2y - 3t = 0$$ we take all terms containing $$t$$ to one side:

$$t\,x - 3t = 2y.$$

Now factor out $$t$$ on the left:

$$t\,(x - 3) = 2y.$$

Provided $$x \neq 3$$, we can solve for $$t$$:

$$t = \dfrac{2y}{x - 3}.$$

Next, substitute this value of $$t$$ into the other straight-line equation $$x - 2t\,y + 3 = 0$$. We obtain:

$$x - 2\left(\dfrac{2y}{x - 3}\right)y + 3 = 0.$$

Simplifying the fraction inside the parentheses gives $$2 \cdot \dfrac{2y}{x - 3}\,y = \dfrac{4y^{2}}{x - 3}$$, so the expression becomes

$$x - \dfrac{4y^{2}}{x - 3} + 3 = 0.$$

To clear the denominator, multiply every term by $$x - 3$$:

$$(x - 3)\,x - 4y^{2} + (x - 3)\,3 = 0.$$

We now expand each product:

$$x^{2} - 3x - 4y^{2} + 3x - 9 = 0.$$

The terms $$-3x$$ and $$+3x$$ cancel, leaving

$$x^{2} - 4y^{2} - 9 = 0.$$

Rearrange to get all constants on the right:

$$x^{2} - 4y^{2} = 9.$$

This is the standard form of a rectangular conic. To see which one, we divide both sides by $$9$$:

$$\dfrac{x^{2}}{9} - \dfrac{4y^{2}}{9} = 1.$$

Write the second fraction as a single square in the denominator:

$$\dfrac{x^{2}}{9} - \dfrac{y^{2}}{\,9/4\,} = 1.$$

In the canonical form $$\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1$$ for a hyperbola, we recognize

$$a^{2} = 9 \;\Longrightarrow\; a = 3, \qquad b^{2} = \dfrac{9}{4} \;\Longrightarrow\; b = \dfrac{3}{2}.$$

For a hyperbola, the conjugate axis has length $$2b$$. Therefore

$$2b = 2\left(\dfrac{3}{2}\right) = 3.$$

Thus the locus is a hyperbola whose conjugate axis is exactly $$3$$ units long.

Hence, the correct answer is Option A.

We are told that the two foci of the required hyperbola are $$(\pm 2,0)$$. For a hyperbola centred at the origin and opening right-left, the standard equation is

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,$$

where

$$c^{2}=a^{2}+b^{2}\quad\text{and}\quad c=\text{distance of each focus from the centre}.$$

Because the foci are at $$(\pm 2,0)$$, we immediately have

$$c=2\quad\Longrightarrow\quad c^{2}=4.$$

Hence, by the relation $$c^{2}=a^{2}+b^{2},$$ we get the first equation

$$a^{2}+b^{2}=4. \quad -(1)$$

The hyperbola also passes through the given point $$P(\sqrt{2},\sqrt{3}).$$ Substituting this point into the general equation gives

$$\frac{(\sqrt{2})^{2}}{a^{2}}-\frac{(\sqrt{3})^{2}}{b^{2}}=1,$$

which simplifies term-wise to

$$\frac{2}{a^{2}}-\frac{3}{b^{2}}=1. \quad -(2)$$

To find the specific values of $$a^{2}$$ and $$b^{2},$$ we solve equations (1) and (2) simultaneously. Let

$$A=a^{2},\qquad B=b^{2}.$$

Then (1) and (2) become

$$A+B=4, \quad -(1')$$

$$\frac{2}{A}-\frac{3}{B}=1. \quad -(2')$$

From (1′), express $$B$$ in terms of $$A$$:

$$B=4-A.$$

Substitute this into (2′):

$$\frac{2}{A}-\frac{3}{\,\,4-A}=1.$$

Take a common denominator $$A(4-A)$$:

$$\frac{2(4-A)-3A}{A(4-A)}=1.$$

Simplify the numerator:

$$\frac{8-2A-3A}{A(4-A)}=1\;\Longrightarrow\;\frac{8-5A}{A(4-A)}=1.$$

Cross-multiply:

$$8-5A=A(4-A).$$

Expand and rearrange everything to one side:

$$8-5A=4A-A^{2}\quad\Longrightarrow\quad 0=4A-A^{2}-8+5A$$

$$\Longrightarrow\quad 0=-A^{2}+9A-8.$$

Multiply by $$-1$$ to get a conventional quadratic:

$$A^{2}-9A+8=0.$$

Factor (or use the quadratic formula):

$$(A-1)(A-8)=0\;\Longrightarrow\;A=1\;\text{or}\;A=8.$$

If $$A=8$$ then $$B=4-8=-4,$$ which is impossible because $$b^{2}$$ must be positive. Hence we accept

$$a^{2}=A=1,\qquad b^{2}=B=4-1=3.$$

Therefore the explicit equation of the hyperbola is

$$\frac{x^{2}}{1}-\frac{y^{2}}{3}=1.$$

Next we need the tangent at the point $$P(\sqrt{2},\sqrt{3}).$$ For a hyperbola of the form $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1,$$ the point-form tangent formula is

$$\frac{xx_{1}}{a^{2}}-\frac{yy_{1}}{b^{2}}=1,$$

where $$(x_{1},y_{1})$$ is the point of tangency. Plugging $$a^{2}=1,\;b^{2}=3,\;x_{1}=\sqrt{2},\;y_{1}=\sqrt{3},$$ we obtain

$$\frac{x\sqrt{2}}{1}-\frac{y\sqrt{3}}{3}=1.$$

To clear the denominator 3, multiply the entire equation by 3:

$$3x\sqrt{2}-y\sqrt{3}=3. \quad -(3)$$

Equation (3) is the straight‐line equation of the required tangent. We now test which of the given options satisfies this linear equation.

Option A: $$(3\sqrt{2},\,2\sqrt{3})$$

 Left-hand side  = $$3(3\sqrt{2})\sqrt{2}-(2\sqrt{3})\sqrt{3}=3\cdot3\cdot2-2\cdot3=18-6=12\neq3.$$ So the point is not on the tangent.

Option B: $$(2\sqrt{2},\,3\sqrt{3})$$

 Left-hand side  = $$3(2\sqrt{2})\sqrt{2}-(3\sqrt{3})\sqrt{3}=3\cdot2\cdot2-3\cdot3=12-9=3,$$ which equals the right-hand side of equation (3). Hence this point does lie on the tangent.

Option C: $$(\sqrt{3},\,\sqrt{2})$$

 Left-hand side  = $$3(\sqrt{3})\sqrt{2}-(\sqrt{2})\sqrt{3}=3\sqrt{6}-\sqrt{6}=2\sqrt{6}\neq3.$$ So the point is not on the tangent.

Option D: $$(-\sqrt{2},\,-\sqrt{3})$$

 Left-hand side  = $$3(-\sqrt{2})\sqrt{2}-(-\sqrt{3})\sqrt{3}=-6+3=-3\neq3.$$ So the point is not on the tangent.

Only Option B satisfies the tangent equation.

Hence, the correct answer is Option B.

Let us consider a standard rectangular hyperbola whose transverse axis is taken along the $$x$$-axis. Its Cartesian equation in the simplest form is written as

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1.$$

For this hyperbola we know the following standard facts:

• The coordinates of the foci are $$(\pm ae,0)$$, so the distance between the two foci is $$2ae.$$

• The length of the conjugate axis (the axis along the $$y$$-direction) is twice the semi-conjugate, that is $$2b.$$

• The relation among the semi-transverse $$a$$, the semi-conjugate $$b$$ and the eccentricity $$e$$ is $$b^{2}=a^{2}(e^{2}-1).$$

Now we translate the verbal condition of the problem into an algebraic equation. We are told that “the length of its conjugate axis is equal to half of the distance between its foci.” That sentence becomes, in symbols,

$$2b=\tfrac12\,(2ae).$$

Simplifying the right-hand side first: $$\tfrac12\,(2ae)=ae.$$ Hence the given condition is

$$2b=ae.$$

We isolate $$b$$ because the formula that connects $$a,b,e$$ has $$b^{2}$$ in it. Dividing both sides by 2 gives

$$b=\frac{ae}{2}.$$

Next we square this result so that we may substitute into the basic identity: $$b^{2}=\left(\frac{ae}{2}\right)^{2}=\frac{a^{2}e^{2}}{4}.$$

But from the standard relation quoted earlier we also have $$b^{2}=a^{2}(e^{2}-1).$$

Because both right-hand sides represent the same $$b^{2},$$ we equate them:

$$\frac{a^{2}e^{2}}{4}=a^{2}(e^{2}-1).$$

Since $$a^{2}$$ is positive and common on both sides, we cancel it to obtain

$$\frac{e^{2}}{4}=e^{2}-1.$$

To clear the fraction we multiply every term by 4, giving

$$e^{2}=4e^{2}-4.$$

Now we bring all terms to one side: $$0=4e^{2}-4-e^{2}=3e^{2}-4.$$

Rearranging, we write

$$3e^{2}=4.$$

Dividing by 3 gives

$$e^{2}=\frac{4}{3}.$$

Taking the positive square root (because the eccentricity of a hyperbola is always >1) yields

$$e=\frac{2}{\sqrt{3}}.$$

We compare this with the answer choices and see that it matches Option A.

Hence, the correct answer is Option A.

We are told that the hyperbola is in its standard (centre at the origin) horizontal form, so its equation can be written as $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1.$$

For this standard form we recall three very important facts:

1. The coordinates of the foci are $$(\pm c,0).$$

2. The relation between the semi-transverse axis $$a$$, the semi-conjugate axis $$b$$ and the focal distance $$c$$ is $$c^2=a^2+b^2.$$

3. The eccentricity is defined as $$e=\dfrac{c}{a},$$ and the corresponding right-hand directrix is $$x=\dfrac{a}{e}.$$

Now we turn to the data given in the question.

First we compute the eccentricity from the quadratic condition $$9e^2-18e+5=0.$$ Dividing every term by $$9$$ gives $$e^2-2e+\dfrac{5}{9}=0.$$ Using the quadratic formula $$e=\dfrac{2\pm\sqrt{(-2)^2-4\cdot1\cdot\dfrac{5}{9}}}{2} =\dfrac{2\pm\sqrt{4-\dfrac{20}{9}}}{2} =\dfrac{2\pm\sqrt{\dfrac{16}{9}}}{2} =\dfrac{2\pm\dfrac{4}{3}}{2}.$$ This produces two numerical values: $$e_1=\dfrac{2+\dfrac{4}{3}}{2}=\dfrac{\dfrac{10}{3}}{2}=\dfrac{5}{3},\qquad e_2=\dfrac{2-\dfrac{4}{3}}{2}=\dfrac{\dfrac{2}{3}}{2}=\dfrac{1}{3}.$$ Because a hyperbola must have $$e>1,$$ we select $$e=\dfrac{5}{3}.$$

The focus supplied is $$S(5,0),$$ so the focal distance is $$c=5.$$

The directrix corresponding to this focus is given as $$5x=9,$$ which is $$x=\dfrac{9}{5}.$$ According to fact 3, for the right-hand side of the hyperbola that directrix must satisfy $$x=\dfrac{a}{e}.$$ Therefore $$\dfrac{a}{e}=\dfrac{9}{5}\quad\Longrightarrow\quad a=\dfrac{9}{5}\,e.$$

Substituting our value $$e=\dfrac{5}{3}$$ we obtain $$a=\dfrac{9}{5}\left(\dfrac{5}{3}\right)=\dfrac{9}{3}=3.$$ Hence $$a^2=3^2=9.$$

With $$a$$ known and the focal distance $$c=5,$$ we use the relation $$c^2=a^2+b^2$$ (fact 2) to find $$b^2:$$ $$25=9+b^2\quad\Longrightarrow\quad b^2=25-9=16.$$

The quantity required in the question is $$a^2-b^2:$$ $$a^2-b^2=9-16=-7.$$

Hence, the correct answer is Option A.

The given hyperbola is $$\frac{x^2}{4} - \frac{y^2}{5} = 1$$. Comparing this with the standard form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, we find $$a^2 = 4$$ and $$b^2 = 5$$, so $$a = 2$$ and $$b = \sqrt{5}$$.

The foci of the hyperbola are at $$(\pm c, 0)$$, where $$c = \sqrt{a^2 + b^2} = \sqrt{4 + 5} = \sqrt{9} = 3$$. Thus, the foci are $$(3, 0)$$ and $$(-3, 0)$$.

We need the extremity of the latus rectum in the first quadrant. The latus rectum through the focus $$(3, 0)$$ is perpendicular to the x-axis, so its equation is $$x = 3$$. Substituting $$x = 3$$ into the hyperbola equation:

$$\frac{(3)^2}{4} - \frac{y^2}{5} = 1 \implies \frac{9}{4} - \frac{y^2}{5} = 1.$$

Rearranging terms:

$$\frac{9}{4} - 1 = \frac{y^2}{5} \implies \frac{5}{4} = \frac{y^2}{5}.$$

Solving for $$y^2$$:

$$y^2 = \frac{5}{4} \times 5 = \frac{25}{4},$$

so $$y = \pm \frac{5}{2}$$. The extremities are $$(3, \frac{5}{2})$$ and $$(3, -\frac{5}{2})$$. The first quadrant point is $$(3, \frac{5}{2})$$.

The tangent to the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ at a point $$(x_1, y_1)$$ is given by $$\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$$. Substituting $$a^2 = 4$$, $$b^2 = 5$$, $$x_1 = 3$$, and $$y_1 = \frac{5}{2}$$:

$$\frac{x \cdot 3}{4} - \frac{y \cdot \frac{5}{2}}{5} = 1.$$

Simplifying the second term:

$$\frac{y \cdot \frac{5}{2}}{5} = \frac{5y}{2} \times \frac{1}{5} = \frac{y}{2},$$

so the equation becomes:

$$\frac{3x}{4} - \frac{y}{2} = 1.$$

Multiplying through by 4 to clear denominators:

$$4 \times \frac{3x}{4} - 4 \times \frac{y}{2} = 4 \times 1 \implies 3x - 2y = 4.$$

Thus, the tangent equation is $$3x - 2y = 4$$.

This tangent meets the x-axis at A (where $$y = 0$$):

$$3x - 2(0) = 4 \implies 3x = 4 \implies x = \frac{4}{3},$$

so A is $$\left(\frac{4}{3}, 0\right)$$.

It meets the y-axis at B (where $$x = 0$$):

$$3(0) - 2y = 4 \implies -2y = 4 \implies y = -2,$$

so B is $$(0, -2)$$.

O is the origin $$(0, 0)$$. Now, compute $$OA^2$$ and $$OB^2$$:

OA is the distance from O to A $$\left(\frac{4}{3}, 0\right)$$, so $$OA = \frac{4}{3}$$ and $$OA^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$$.

OB is the distance from O to B $$(0, -2)$$, so $$OB = 2$$ and $$OB^2 = 2^2 = 4$$.

Therefore,

$$OA^2 - OB^2 = \frac{16}{9} - 4 = \frac{16}{9} - \frac{36}{9} = -\frac{20}{9}.$$

Hence, the correct answer is Option A.

We have the rectangular hyperbola in standard form

$$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$$

whose semi-transverse axis is $$a=3$$ and semi-conjugate axis is $$b=2$$. For this curve the standard parametric equations are

$$x=a\sec\alpha,\qquad y=b\tan\alpha.$$

Hence the points

$$P\;(3\sec\theta,\,2\tan\theta),\qquad Q\;(3\sec\phi,\,2\tan\phi)$$

indeed lie on the hyperbola. It is given that

$$\theta+\phi=\frac{\pi}{2}.$$

To find the intersection of the normals at these two points, we first write the slope of the normal at a general point.

The hyperbola is represented by $$F(x,y)=\frac{x^{2}}{9}-\frac{y^{2}}{4}-1=0.$$ Differentiating implicitly with respect to $$x$$,

$$\frac{2x}{9}-\frac{2y}{4}\,\frac{dy}{dx}=0 \;\;\Longrightarrow\;\;\frac{dy}{dx}=\frac{4x}{9y}.$$

Thus the slope of the tangent at $$(x,y)$$ is $$m_t=\dfrac{4x}{9y}$$, and the slope of the normal is the negative reciprocal:

$$m_n=-\frac{1}{m_t}=-\frac{9y}{4x}.$$

Applying this to the two parametric points:

For $$P(3\sec\theta,2\tan\theta):$$

$$m_P=-\frac{9(2\tan\theta)}{4(3\sec\theta)} =-\frac{18\tan\theta}{12\sec\theta} =-\frac{3\tan\theta}{2\sec\theta} =-\frac{3\sin\theta}{2}.$$

For $$Q(3\sec\phi,2\tan\phi):$$

$$m_Q=-\frac{9(2\tan\phi)}{4(3\sec\phi)} =-\frac{3\tan\phi}{2\sec\phi} =-\frac{3\sin\phi}{2}.$$

Now we write the equations of the two normals.

Through $$P(3\sec\theta,2\tan\theta):$$

$$y-2\tan\theta=m_P\bigl(x-3\sec\theta\bigr).$$

Through $$Q(3\sec\phi,2\tan\phi):$$

$$y-2\tan\phi=m_Q\bigl(x-3\sec\phi\bigr).$$

Re-arranging each into the slope-intercept form $$y=mx+c$$:

For $$P$$ (putting $$m_1=m_P$$): $$$ y=m_1x+\underbrace{\bigl(2\tan\theta-3m_1\sec\theta\bigr)}_{c_1}. $$$

For $$Q$$ (putting $$m_2=m_Q$$): $$$ y=m_2x+\underbrace{\bigl(2\tan\phi-3m_2\sec\phi\bigr)}_{c_2}. $$$

Let us evaluate the two intercepts explicitly.

Because $$m_1=-\dfrac{3\sin\theta}{2}$$,

$$$ c_1=2\tan\theta-3m_1\sec\theta =2\frac{\sin\theta}{\cos\theta} -3\Bigl(-\frac{3\sin\theta}{2}\Bigr)\frac{1}{\cos\theta} =\frac{4\sin\theta}{\cos\theta}+\frac{9\sin\theta}{2\cos\theta} =\frac{13\sin\theta}{2\cos\theta} =\frac{13}{2}\tan\theta. $$$

Similarly, with $$m_2=-\dfrac{3\sin\phi}{2}$$,

$$$ c_2=2\tan\phi-3m_2\sec\phi =\frac{13}{2}\tan\phi. $$$

The intersection of the two straight lines

$$y=m_1x+c_1,\qquad y=m_2x+c_2$$

has ordinates obtained from the determinant formula

$$Y=\frac{m_1c_2-m_2c_1}{m_1-m_2}.$$

Substituting the explicit expressions,

$$$ Y=\frac{\displaystyle m_1\Bigl(\frac{13}{2}\tan\phi\Bigr) -m_2\Bigl(\frac{13}{2}\tan\theta\Bigr)} {m_1-m_2} =\frac{13}{2}\; \frac{m_1\tan\phi-m_2\tan\theta}{m_1-m_2}. $$$

Insert the values $$m_1=-\dfrac{3\sin\theta}{2},\; m_2=-\dfrac{3\sin\phi}{2}$$:

$$$ m_1\tan\phi=-\frac{3\sin\theta}{2}\,\frac{\sin\phi}{\cos\phi} =-\frac{3\sin\theta\sin\phi}{2\cos\phi}, $$$

$$$ m_2\tan\theta=-\frac{3\sin\phi}{2}\,\frac{\sin\theta}{\cos\theta} =-\frac{3\sin\phi\sin\theta}{2\cos\theta}. $$$

Therefore

$$$ m_1\tan\phi-m_2\tan\theta =-\frac{3\sin\theta\sin\phi}{2}\Bigl(\frac{1}{\cos\phi}-\frac{1}{\cos\theta}\Bigr). $$$

And

$$$ m_1-m_2=-\frac{3\sin\theta}{2}+\frac{3\sin\phi}{2} =\frac{3}{2}\bigl(\sin\phi-\sin\theta\bigr). $$$

Substituting back,

$$$ Y=\frac{13}{2}\; \frac{\displaystyle -\dfrac{3\sin\theta\sin\phi}{2} \Bigl(\dfrac{1}{\cos\phi}-\dfrac{1}{\cos\theta}\Bigr)} {\displaystyle \dfrac{3}{2}\,(\sin\phi-\sin\theta)} =-\frac{13}{2}\, \frac{\sin\theta\sin\phi\;\bigl(\frac{1}{\cos\phi}-\frac{1}{\cos\theta}\bigr)} {\cos\theta\cos\phi\;(\, \frac{\sin\phi-\sin\theta}{\cos\theta\cos\phi}\,)} =-\frac{13}{2}\, \frac{\sin\theta\sin\phi\;(\cos\theta-\cos\phi)} {\cos\theta\cos\phi\;(\sin\phi-\sin\theta)}. $$$

At this point we employ the crucial relation $$\theta+\phi=\dfrac{\pi}{2}$$, which implies

$$\sin\phi=\cos\theta,\qquad \cos\phi=\sin\theta.$$

Putting these into the last expression, every factor cancels:

$$$ Y=-\frac{13}{2}\, \frac{(\sin\theta)(\cos\theta)\;(\cos\theta-\sin\theta)} {(\cos\theta)(\sin\theta)\;(\cos\theta-\sin\theta)} =-\frac{13}{2}. $$$

Thus the ordinate (the $$y$$-coordinate) of the point where the two normals meet is

$$Y=-\frac{13}{2}.$$

Hence, the correct answer is Option D.

We start with the hyperbola equation: $$\frac{x^2}{4} - \frac{y^2}{2} = 1$$. Here, $$a^2 = 4$$ and $$b^2 = 2$$, so $$a = 2$$ and $$b = \sqrt{2}$$. The equation of a tangent to this hyperbola with slope $$m$$ is given by: $$y = mx \pm \sqrt{a^2 m^2 - b^2} = mx \pm \sqrt{4m^2 - 2}$$.

Let the tangent be $$y = mx + c$$, where $$c = \pm \sqrt{4m^2 - 2}$$. This tangent meets the x-axis at point P and the y-axis at point Q.

When the tangent meets the x-axis, $$y = 0$$: $$0 = mx + c \implies x = -\frac{c}{m}$$ So, P has coordinates $$\left(-\frac{c}{m}, 0\right)$$.

When the tangent meets the y-axis, $$x = 0$$: $$y = m \cdot 0 + c = c$$ So, Q has coordinates $$(0, c)$$.

We are given that O is the origin $$(0, 0)$$, and OPRQ is a rectangle. The diagonals of a rectangle bisect each other. The diagonals are OR (from O to R) and PQ (from P to Q). Let R be $$(x, y)$$.

The midpoint of OR is $$\left(\frac{x}{2}, \frac{y}{2}\right)$$.

The midpoint of PQ, where P is $$\left(-\frac{c}{m}, 0\right)$$ and Q is $$(0, c)$$, is $$\left(\frac{-\frac{c}{m} + 0}{2}, \frac{0 + c}{2}\right) = \left(-\frac{c}{2m}, \frac{c}{2}\right)$$.

Setting the midpoints equal: $$\frac{x}{2} = -\frac{c}{2m} \quad \text{and} \quad \frac{y}{2} = \frac{c}{2}$$ From the second equation: $$\frac{y}{2} = \frac{c}{2} \implies y = c$$.

From the first equation: $$\frac{x}{2} = -\frac{c}{2m} \implies x = -\frac{c}{m}$$.

Substituting $$c = y$$ into $$x = -\frac{c}{m}$$ gives $$x = -\frac{y}{m}$$, so $$m = -\frac{y}{x}$$.

We also know $$c = \pm \sqrt{4m^2 - 2}$$. Substituting $$c = y$$ and $$m = -\frac{y}{x}$$: $$y = \pm \sqrt{4\left(-\frac{y}{x}\right)^2 - 2} = \pm \sqrt{4 \cdot \frac{y^2}{x^2} - 2} = \pm \sqrt{\frac{4y^2}{x^2} - 2}$$

Squaring both sides to eliminate the square root: $$y^2 = \left(\pm \sqrt{\frac{4y^2}{x^2} - 2}\right)^2 = \frac{4y^2}{x^2} - 2$$

Rearranging terms: $$y^2 - \frac{4y^2}{x^2} = -2$$ Multiplying both sides by $$x^2$$ to clear the denominator: $$y^2 x^2 - 4y^2 = -2x^2$$ Bringing all terms to one side: $$x^2 y^2 - 4y^2 + 2x^2 = 0$$

Assuming $$x \neq 0$$ and $$y \neq 0$$ (since R is not on the axes), divide both sides by $$x^2 y^2$$: $$\frac{x^2 y^2}{x^2 y^2} - \frac{4y^2}{x^2 y^2} + \frac{2x^2}{x^2 y^2} = 0 \implies 1 - \frac{4}{x^2} + \frac{2}{y^2} = 0$$ Rearranging: $$\frac{4}{x^2} - \frac{2}{y^2} = 1$$

Comparing with the options, this matches option D. Hence, the correct answer is Option D.

The given line is $$y=\alpha x+\beta$$ and the given conic is the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$.

Step 1 Substitute the line in the hyperbola.
Putting $$y=\alpha x+\beta$$ in the hyperbola, we get

$$\frac{x^{2}}{a^{2}}-\frac{(\alpha x+\beta)^{2}}{b^{2}}=1.$$

Multiply by $$a^{2}b^{2}$$ to clear denominators:

$$b^{2}x^{2}-a^{2}(\alpha x+\beta)^{2}-a^{2}b^{2}=0.$$

Step 2 Write the resulting quadratic in $$x$$.
Expand and collect terms of powers of $$x$$:

$$\bigl(b^{2}-a^{2}\alpha^{2}\bigr)x^{2}-2a^{2}\alpha\beta\,x-a^{2}\bigl(\beta^{2}+b^{2}\bigr)=0.$$

This is a quadratic of the form $$A x^{2}+B x+C=0$$ with
$$A=b^{2}-a^{2}\alpha^{2},\quad B=-2a^{2}\alpha\beta,\quad C=-a^{2}\bigl(\beta^{2}+b^{2}\bigr).$$

Step 3 Impose the condition for tangency.
A line is tangent to a conic when this quadratic has equal roots, i.e. its discriminant is zero:

$$B^{2}-4AC=0.$$

Substitute $$A,B,C$$:

$$\bigl(-2a^{2}\alpha\beta\bigr)^{2}-4\bigl(b^{2}-a^{2}\alpha^{2}\bigr)\!\bigl(-a^{2}(\beta^{2}+b^{2})\bigr)=0.$$

Simplify (divide by the common factor $$4$$ first):

$$a^{4}\alpha^{2}\beta^{2}+a^{2}\bigl(b^{2}-a^{2}\alpha^{2}\bigr)\bigl(\beta^{2}+b^{2}\bigr)=0.$$

Expand the second product:

$$a^{4}\alpha^{2}\beta^{2}+a^{2}\bigl(b^{2}-a^{2}\alpha^{2}\bigr)\beta^{2}+a^{2}\bigl(b^{2}-a^{2}\alpha^{2}\bigr)b^{2}=0.$$

Group the terms containing $$\beta^{2}$$:

$$\beta^{2}\,a^{2}\Bigl(a^{2}\alpha^{2}+b^{2}-a^{2}\alpha^{2}\Bigr)+a^{2}b^{2}\bigl(b^{2}-a^{2}\alpha^{2}\bigr)=0.$$

Inside the parenthesis $$\bigl(a^{2}\alpha^{2}+b^{2}-a^{2}\alpha^{2}\bigr)=b^{2}$$, therefore

$$a^{2}b^{2}\beta^{2}+a^{2}b^{2}\bigl(b^{2}-a^{2}\alpha^{2}\bigr)=0.$$

Divide both sides by the non-zero factor $$a^{2}b^{2}$$:

$$\beta^{2}+b^{2}-a^{2}\alpha^{2}=0.$$

Step 4 Obtain the locus equation.
Re-arrange:

$$a^{2}\alpha^{2}-\beta^{2}=b^{2}.$$

Divide by $$b^{2}$$ to see the standard form:

$$\frac{a^{2}\alpha^{2}}{b^{2}}-\frac{\beta^{2}}{b^{2}}=1$$
or equivalently
$$\frac{\alpha^{2}}{\,b^{2}/a^{2}\,}-\frac{\beta^{2}}{b^{2}}=1.$$

Step 5 Identify the curve.
This equation has a positive $$\alpha^{2}$$ term and a negative $$\beta^{2}$$ term, matching the general form $$\frac{X^{2}}{A^{2}}-\frac{Y^{2}}{B^{2}}=1$$ of a hyperbola. Hence the locus of $$P(\alpha,\beta)$$ is a hyperbola.

Option D which is: a hyperbola

For the ellipse

$$\frac{x^2}{18}+\frac{y^2}{12}=1$$

we have

$$a^2=18,\qquad b^2=12$$

Hence,

$$c^2=a^2-b^2=6$$

and the eccentricity is

$$e=\frac{c}{a}=\frac{\sqrt6}{3\sqrt2}=\frac1{\sqrt3}$$

Therefore, the hyperbola has eccentricity

$$e_H=\frac1e=\sqrt3$$

and the same foci, so

$$c^2=6$$

For the hyperbola,

$$e_H=\frac{c}{a_H}=\sqrt3$$

Hence,

$$a_H^2=\frac{c^2}{e_H^2}=\frac6{3}=2$$

Also,

$$b_H^2=c^2-a_H^2=6-2=4$$

Therefore, the equation of the hyperbola is

$$\frac{x^2}{2}-\frac{y^2}{4}=1$$

The parabola is

$$\sqrt5,y=x^2$$

or

$$y=\frac{x^2}{\sqrt5}$$

Substituting into the hyperbola,

$$\frac{x^2}{2}-\frac1{4}\left(\frac{x^2}{\sqrt5}\right)^2=1$$

$$\frac{x^2}{2}-\frac{x^4}{20}=1$$

Multiplying by $$20$$,

$$10x^2-x^4=20$$

$$x^4-10x^2+20=0$$

Let

$$t=x^2$$

Then

$$t^2-10t+20=0$$

$$t=5\pm\sqrt5$$

Hence the two points are

$$P\left(\sqrt{5-\sqrt5},\frac{5-\sqrt5}{\sqrt5}\right)$$

and

$$Q\left(\sqrt{5+\sqrt5},\frac{5+\sqrt5}{\sqrt5}\right)$$

Now,

$$d^2=(x_2-x_1)^2+(y_2-y_1)^2$$

First,

$$y_2-y_1=\frac{(5+\sqrt5)-(5-\sqrt5)}{\sqrt5}=2$$

Also,

$$x_2^2+x_1^2=10$$

and

$$x_1x_2=\sqrt{(5-\sqrt5)(5+\sqrt5)}=\sqrt{20}=2\sqrt5$$

Therefore,

$$ (x_2-x_1)^2=x_2^2+x_1^2-2x_1x_2 $$

$$=10-4\sqrt5$$

Hence,

$$d^2=(10-4\sqrt5)+4$$

$$=14-4\sqrt5$$

Thus,

$$a=14,\qquad b=-4$$

and

$$a-b=14-(-4)$$

$$=18$$

Hence,

$$\boxed{18}$$

$$\frac{2b^{2}}{a}=8 \;\Longrightarrow\; \frac{b^{2}}{a}=4 \;\Longrightarrow\; b^{2}=4a \;$$ $$-(1)$$

The point $$P(10,\,2\sqrt{15})$$ lies on the hyperbola, so it satisfies the equation:
$$\frac{10^{2}}{a^{2}}-\frac{(2\sqrt{15})^{2}}{b^{2}} = 1$$
$$\Longrightarrow \frac{100}{a^{2}}-\frac{60}{b^{2}} = 1$$ $$-(2)$$

From $$(1)$$: $$a = \frac{b^{2}}{4} \;\Longrightarrow\; a^{2}=\frac{b^{4}}{16}$$.

Substituting $$a^{2}$$ from above and $$b^{2}$$ itself into $$(2)$$:
$$\frac{100}{\,b^{4}/16\,} - \frac{60}{b^{2}} = 1$$
$$\Longrightarrow \frac{1600}{b^{4}} - \frac{60}{b^{2}} - 1 = 0$$.

Put $$t = \frac{1}{b^{2}}$$. The quadratic becomes
$$1600t^{2}-60t-1=0$$.

Solving, the discriminant is $$\Delta = (-60)^{2}-4\cdot1600(-1)=10000$$, hence
$$t = \frac{60\pm100}{3200}\;.$$

Only the positive root is acceptable: $$t = \frac{160}{3200}=\frac{1}{20}$$, so
$$b^{2} = 20$$.

From $$(1)$$, $$a = \frac{b^{2}}{4} = \frac{20}{4}=5$$.

Hence $$c^{2}=a^{2}+b^{2}=25+20=45 \;\Longrightarrow\; c=\sqrt{45}=3\sqrt{5}$$.

The base $$SS'$$ of $$\triangle PSS'$$ lies on the x-axis, so its length is $$2c$$.
The perpendicular height from $$P$$ to this base equals the y-coordinate of $$P$$, namely $$2\sqrt{15}$$.

 the area of the triangle is
$$\text{Area}= \frac12 \times (2c)\times(2\sqrt{15}) = 2c\sqrt{15}$$.
$$\bigl(\text{Area}\bigr)^{2}= \bigl(2c\sqrt{15}\bigr)^{2}=4c^{2}\times15 = 60c^{2}$$.

Using $$c^{2}=45$$ we get
$$60c^{2}=60\times45=2700$$.

$$9x - x^{2} - 13 \gt 0$$
$$\log_{7}(9x - x^{2} - 13) \gt 1 \;\;\Longrightarrow\;\; 9x - x^{2} - 13 \gt 7$$

The second inequality is stronger, so we keep it:
$$9x - x^{2} - 13 \gt 7 \;\;\Longrightarrow\;\; 9x - x^{2} - 20 \gt 0$$
$$-x^{2} + 9x - 20 \gt 0 \;\;\Longrightarrow\;\; x^{2} - 9x + 20 \lt 0$$
$$x^{2} - 9x + 20 = (x-4)(x-5)$$

 the domain is the open interval$$(4,\,5)$$
Thus $$m = 4$$ and $$n = 5$$.

For the hyperbola $$\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1$$:

• Eccentricity $$e = \dfrac{n}{3} = \dfrac{5}{3}$$
• Length of the latus rectum $$L = \dfrac{8m}{3} = \dfrac{8 \times 4}{3} = \dfrac{32}{3}$$

The standard results for such a hyperbola are

$$e^{2} = 1 + \dfrac{b^{2}}{a^{2}}$$
$$L = \dfrac{2b^{2}}{a}$$

$$\left(\dfrac{5}{3}\right)^{2} = 1 + \dfrac{b^{2}}{a^{2}} \;\;\Longrightarrow\;\; \dfrac{25}{9} = 1 + \dfrac{b^{2}}{a^{2}}$$

$$\dfrac{b^{2}}{a^{2}} = \dfrac{25}{9} - 1 = \dfrac{16}{9} \;\;\Longrightarrow\;\; b^{2} = \dfrac{16}{9}\,a^{2} \;$$

Using the latus-rectum length:

$$\dfrac{2b^{2}}{a} = \dfrac{32}{3} \;\;\Longrightarrow\;\; b^{2} = \dfrac{16}{3}\,a \;$$

$$\dfrac{16}{9}\,a^{2} = \dfrac{16}{3}\,a \;\;\Longrightarrow\;\; \dfrac{1}{9}a^{2} = \dfrac{1}{3}a \;\;\Longrightarrow\;\; a^{2} = 3a \;\;\Longrightarrow\;\; a(a-3)=0$$

Since $$a\gt0$$, we take $$a = 3$$, giving $$a^{2} = 9$$.

Substitute back to find $$b^{2}$$:

$$b^{2} = \dfrac{16}{9}\,a^{2} = \dfrac{16}{9}\times9 = 16$$

$$b^{2} - a^{2} = 16 - 9 = 7$$

The hyperbola is $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$, with foci $$F_{1}(-c,0)$$ and $$F_{2}(c,0)$$, where $$c^{2}=a^{2}+b^{2}$$.

For any point on this hyperbola, the difference of its focal distances equals the length of the transverse axis: $$|PF_{2}-PF_{1}|=2a$$.

Let
$$d_{1}=PF_{1},\qquad d_{2}=PF_{2}$$
for the point $$P(4,3)$$. The data give the following:

1. Sum of focal distances
$$d_{1}+d_{2}=8\sqrt{\frac{5}{3}}$$ $$-(1)$$

2. Using coordinates,
$$d_{1}^{2}=(4-c)^{2}+3^{2}=25-8c+c^{2}$$
$$d_{2}^{2}=(4+c)^{2}+3^{2}=25+8c+c^{2}$$

Hence
$$d_{2}^{2}-d_{1}^{2}=16c$$ $$-(2)$$

3. From the property of a hyperbola,
$$d_{2}-d_{1}=2a$$ $$-(3)$$

Using $$(1)$$ and $$(3)$$,
$$d_{2}=\frac{(d_{1}+d_{2})+(d_{2}-d_{1})}{2}=\frac{S}{2}+a,\quad d_{1}=\frac{S}{2}-a,$$ where $$S=d_{1}+d_{2}=8\sqrt{\frac{5}{3}}$$.

The product $$(d_{2}+d_{1})(d_{2}-d_{1})$$ equals $$16c$$ by $$(2)$$, so $$(d_{2}+d_{1})(d_{2}-d_{1})=S\,(2a)=16c.$$ Thus $$c=\frac{aS}{8}.$$

Squaring and substituting $$S^{2}=\left(8\sqrt{\frac{5}{3}}\right)^{2}=\frac{320}{3},$$ $$c^{2}=\frac{a^{2}S^{2}}{64}=a^{2}\,\frac{5}{3}.$$ But $$c^{2}=a^{2}+b^{2},$$ therefore $$a^{2}+b^{2}=a^{2}\,\frac{5}{3}\;\;\Longrightarrow\;\;b^{2}=\frac{2}{3}a^{2}.$$ $$-(4)$$

The point $$P(4,3)$$ lies on the hyperbola, so $$\frac{4^{2}}{a^{2}}-\frac{3^{2}}{b^{2}}=1.$$ Using $$(4)$$, $$\frac{16}{a^{2}}-\frac{9}{(2/3)a^{2}}=1 \;\;\Longrightarrow\;\; \frac{16}{a^{2}}-\frac{27}{2a^{2}}=1 \;\;\Longrightarrow\;\; \frac{2.5}{a^{2}}=1.$$ Hence $$a^{2}=\frac{5}{2},\qquad b^{2}=\frac{2}{3}\cdot\frac{5}{2}=\frac{5}{3},\qquad c^{2}=a^{2}+b^{2}=\frac{25}{6}.$$

Length of the latus rectum
For a hyperbola, $$l=\frac{2b^{2}}{a}.$$ Thus $$l=\frac{2(\frac{5}{3})}{\sqrt{\frac{5}{2}}}=\frac{10}{3}\sqrt{\frac{2}{5}},\qquad l^{2}=\frac{40}{9}.$$ Therefore $$9l^{2}=9\cdot\frac{40}{9}=40.$$

Product of the focal distances
Using $$S^{2}=d_{1}^{2}+d_{2}^{2}+2d_{1}d_{2},$$ first note $$d_{1}^{2}+d_{2}^{2}=(25-8c+c^{2})+(25+8c+c^{2})=50+2c^{2}.$$ Hence $$S^{2}=50+2c^{2}+2m \;\;\Longrightarrow\;\; m=\frac{S^{2}-50-2c^{2}}{2}.$$ Substituting $$S^{2}=\frac{320}{3},\;c^{2}=\frac{25}{6},$$ $$m=\frac{\frac{320}{3}-50-\frac{25}{3}}{2} =\frac{\frac{320-150-25}{3}}{2} =\frac{\frac{145}{3}}{2} =\frac{145}{6}.$$ Therefore $$6m=145.$$

Required value
$$9l^{2}+6m=40+145=185.$$

Hence the correct option is Option C (185).

$$S(1, 14), \quad S'(1, -12), \quad P(1, 6)$$

$$SP = \vert{}14 - 6\vert{} = 8$$

$$S'P = \vert{}6 - (-12)\vert{} = 18$$

$$2b = \vert{}SP - S'P\vert{} = \vert{}8 - 18\vert{} = 10 \implies b = 5$$

$$2be = SS' = 14 - (-12) = 26 \implies be = 13$$

$$a^2 = b^2e^2 - b^2 = (be)^2 - b^2 = 13^2 - 5^2 = 169 - 25 = 144$$

$$\text{L.R.} = \frac{2a^2}{b} = \frac{2(144)}{5} = \frac{288}{5}$$

The hyperbola is $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ with centre at the origin and transverse axis along the $$x$$-axis.

Step 1: Use the distance between the foci.
For this hyperbola the foci are $$\bigl(\pm c,0\bigr)$$ where $$c^{2}=a^{2}+b^{2}$$.
Given distance between the foci $$=2c=6 \; \Longrightarrow \; c=3.$$p>

Step 2: Use the distance between the directrices.
For this hyperbola the directrices are $$x=\pm\dfrac{a}{e}$$ where the eccentricity $$e=\dfrac{c}{a}$$.
Hence the distance between the directrices is $$2\left(\dfrac{a}{e}\right)=\dfrac{2a^{2}}{c}.$$ Given distance $$=\dfrac{8}{3}$$, so $$\dfrac{2a^{2}}{c}=\dfrac{8}{3} \quad\Longrightarrow\quad \dfrac{2a^{2}}{3}=\dfrac{8}{3} \quad\Longrightarrow\quad 2a^{2}=8 \quad\Longrightarrow\quad a^{2}=4.$$

Step 3: Find $$b^{2}$$.
Using $$c^{2}=a^{2}+b^{2}$$ gives $$9=4+b^{2}\quad\Longrightarrow\quad b^{2}=5.$$ Thus the hyperbola is $$\dfrac{x^{2}}{4}-\dfrac{y^{2}}{5}=1.$$

Step 4: Coordinates of the intersection points with $$x=\alpha$$.
Substitute $$x=\alpha$$ in the hyperbola: $$\dfrac{\alpha^{2}}{4}-\dfrac{y^{2}}{5}=1 \;\Longrightarrow\; \dfrac{y^{2}}{5}=\dfrac{\alpha^{2}}{4}-1 \;\Longrightarrow\; y^{2}=5\left(\dfrac{\alpha^{2}}{4}-1\right) =\dfrac{5\alpha^{2}}{4}-5.$$ Hence $$A\bigl(\alpha,+\sqrt{\tfrac{5\alpha^{2}}{4}-5}\bigr),\; B\bigl(\alpha,-\sqrt{\tfrac{5\alpha^{2}}{4}-5}\bigr).$$

Step 5: Area of $$\triangle AOB$$.
Base $$AB$$ is a vertical segment of length $$2\sqrt{\tfrac{5\alpha^{2}}{4}-5}.$$ The perpendicular distance from the origin $$O(0,0)$$ to the line $$x=\alpha$$ is $$|\alpha|$$.
Therefore $$\text{Area}= \dfrac12\,( \text{base} )\,( \text{height} ) =\dfrac12\left[2\sqrt{\tfrac{5\alpha^{2}}{4}-5}\right]|\alpha| =|\alpha|\sqrt{\tfrac{5\alpha^{2}}{4}-5}.$$

The given area is $$4\sqrt{15}$$, so (taking $$\alpha\gt 0$$ because only $$\alpha^{2}$$ is needed) $$\alpha\sqrt{\tfrac{5\alpha^{2}}{4}-5}=4\sqrt{15}.$$ Square both sides: $$\alpha^{2}\left(\dfrac{5\alpha^{2}}{4}-5\right)=240.$$ Put $$\alpha^{2}=X$$: $$X\left(\dfrac{5X-20}{4}\right)=240 \;\Longrightarrow\; X(5X-20)=960 \;\Longrightarrow\; 5X^{2}-20X-960=0 \;\Longrightarrow\; X^{2}-4X-192=0.$$ Solve the quadratic: $$X=\dfrac{4\pm\sqrt{4^{2}+4\cdot192}}{2} =\dfrac{4\pm28}{2}\;\Longrightarrow\; X=16\; (\text{or }-12).$$ Since $$X=\alpha^{2}\gt 0$$, we take $$X=16.$$

Result.
$$\alpha^{2}=16.$$

Option B which is: $$16$$

The two foci are given as $$(3,5)$$ and $$(3,-4)$$. Since the $$x$$-coordinates are equal, the transverse axis of the hyperbola is vertical and the centre is the midpoint of the foci.

Centre $$\,(h,k)=\left(3,\dfrac{5+(-4)}{2}\right)=(3,0.5)$$.

The distance between the foci equals $$2c$$. Hence
$$2c=\sqrt{(3-3)^2+(5-(-4))^2}=|5-(-4)|=9 \;\Longrightarrow\; c=\dfrac{9}{2}=4.5$$.

The eccentricity $$e$$ satisfies the quadratic equation $$6e^2-11e+3=0$$.
Solving, $$e=\dfrac{11\pm\sqrt{11^2-4\cdot6\cdot3}}{12}= \dfrac{11\pm7}{12}$$, giving $$e_1=\dfrac{18}{12}=\dfrac32$$ and $$e_2=\dfrac{4}{12}=\dfrac13$$.

For a hyperbola $$e\gt1$$, so we take $$e=\dfrac32$$.

For a hyperbola $$e=\dfrac{c}{a}\; \Longrightarrow\; a=\dfrac{c}{e}= \dfrac{4.5}{1.5}=3 \quad\Rightarrow\quad a^2=9$$.

The relation between the semi-axes is $$c^2=a^2+b^2$$. Hence
$$b^2=c^2-a^2=(4.5)^2-9=20.25-9=11.25=\dfrac{45}{4}$$.

The length of the latus rectum of a hyperbola is $$L=\dfrac{2b^2}{a}$$.
$$L=\dfrac{2\left(\dfrac{45}{4}\right)}{3}= \dfrac{90}{4}\times\dfrac{1}{3}= \dfrac{90}{12}=7.5=\dfrac{15}{2}$$.

Therefore, the required length of the latus rectum is $$\dfrac{15}{2}$$.

Option C which is: $$\frac{15}{2}$$

$$e^2 = 1 + \frac{b^2}{a^2} \implies 3 = 1 + \frac{b^2}{4} \implies \frac{b^2}{4} = 2 \implies b^2 = 8$$.

Equation: $$\frac{x^2}{4} - \frac{y^2}{8} = 1$$. 
Equilateral Triangle Property: In $$\triangle OPQ$$, if $$P = (x_1, y_1)$$, then because it's equilateral and $$PQ$$ is vertical, the angle $$POX$$ is $$30^\circ$$.

Thus, $$\tan 30^\circ = \frac{y_1}{x_1} \implies \frac{1}{\sqrt{3}} = \frac{y_1}{x_1} \implies y_1 = \frac{x_1}{\sqrt{3}}$$.

 Solve for Point P:

Substitute $$y^2 = \frac{x^2}{3}$$ into the hyperbola:

$$\frac{x^2}{4} - \frac{x^2/3}{8} = 1 \implies \frac{x^2}{4} - \frac{x^2}{24} = 1 \implies \frac{6x^2 - x^2}{24} = 1 \implies 5x^2 = 24 \implies x^2 = \frac{24}{5}$$.
Area of Triangle:

Area $$= \frac{\sqrt{3}}{4} (\text{side})^2$$. Side $$PQ = 2y_1$$.

Side length squared $$s^2 = (2y_1)^2 = 4 \cdot \frac{x^2}{3} = 4 \cdot \frac{24/5}{3} = \frac{32}{5}$$.

Area $$= \frac{\sqrt{3}}{4} \cdot \frac{32}{5} = \frac{8\sqrt{3}}{5}$$.

Correct Answer: B ($$8\sqrt{3}/5$$)

Let the equation of the ellipse be $$\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$$.

Eccentricity $$e_E = \frac{\sqrt{3}}{2}$$

Equation of directrices: $$x = \pm \frac{A}{e_E} = \pm \frac{4\sqrt{6}}{3}$$

$$\frac{A}{\left(\frac{\sqrt{3}}{2}\right)} = \frac{4\sqrt{6}}{3} \implies \frac{2A}{\sqrt{3}} = \frac{4\sqrt{6}}{3}$$

$$A = \frac{4\sqrt{18}}{6} = \frac{12\sqrt{2}}{6} = 2\sqrt{2}$$

$$B^2 = A^2(1 - e_E^2)$$: $$A^2 = (2\sqrt{2})^2 = 8$$

$$B^2 = 8\left(1 - \frac{3}{4}\right) = 8\left(\frac{1}{4}\right) = 2 \implies B = \sqrt{2}$$

Length of semi-major axis of $$E = A = 2\sqrt{2}$$

Length of minor axis of $$E = 2B = 2\sqrt{2}$$

The hyperbola is $$H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with eccentricity $$e_H$$.

$$e_H = \text{length of semi-major axis of } E$$: $$e_H = 2\sqrt{2}$$

latus rectum of $$H = \text{length of minor axis of } E$$: $$\frac{2b^2}{a} = 2\sqrt{2} \implies b^2 = a\sqrt{2} \quad \text{--- (1)}$$

Using $$b^2 = a^2(e_H^2 - 1)$$: $$b^2 = a^2\big((2\sqrt{2})^2 - 1\big) = a^2(8 - 1) = 7a^2 \quad \text{--- (2)}$$

$$7a^2 = a\sqrt{2}$$

$$7a = \sqrt{2} \implies a = \frac{\sqrt{2}}{7}$$

$$b^2 = 7\left(\frac{\sqrt{2}}{7}\right)^2 = 7\left(\frac{2}{49}\right) = \frac{2}{7}$$

$$\text{Distance} = 2 \times \left(\frac{\sqrt{2}}{7}\right) \times 2\sqrt{2} = \frac{4 \times 2}{7} = \frac{8}{7}$$

Standard form of Hyperbola: $$\frac{x^2}{9} - \frac{y^2}{1} = 1$$ (where $$a=3, b=1$$).

A line $$y = mx + c$$ does not meet a hyperbola if it passes between the two branches or doesn't touch them. Specifically, for the line to not intersect, $$c^2 > a^2m^2 - b^2$$ must be false (it must intersect) OR we look at the slopes.

Rewrite the line: $$2y = -\alpha x + 1 \implies y = (-\frac{\alpha}{2})x + \frac{1}{2}$$.

Here $$m = -\frac{\alpha}{2}$$ and $$c = \frac{1}{2}$$.

The condition for a line to be a tangent is $$c^2 = a^2m^2 - b^2$$.

For it to not meet, we need $$c^2 < a^2m^2 - b^2$$.

$$(\frac{1}{2})^2 < (3^2)(-\frac{\alpha}{2})^2 - 1^2$$

$$\frac{1}{4} < 9(\frac{\alpha^2}{4}) - 1 \implies \frac{5}{4} < \frac{9\alpha^2}{4} \implies \alpha^2 > \frac{5}{9} \approx 0.555$$

Evaluate options:

o A: $$0.6^2 = 0.36$$ (False)

o B: $$0.5^2 = 0.25$$ (False)

o C: $$0.7^2 = 0.49$$ (False)

o D: $$0.8^2 = 0.64$$ (True, as $$0.64 > 0.555$$)

Correct Option: D

We start by considering the ellipse $$\frac{x^2}{144} + \frac{y^2}{169} = 1$$. Here $$a^2 = 169$$ (since $$b^2 = 169 > 144 = a^2$$, the major axis is along the y-axis), and $$b^2 = 144$$.

Next, consider the hyperbola $$\frac{x^2}{16} - \frac{y^2}{\lambda^2} = -1$$, which can be written as $$\frac{y^2}{\lambda^2} - \frac{x^2}{16} = 1$$, indicating a conjugate hyperbola with transverse axis along the y-axis.

We find the foci of the ellipse: for the ellipse with major axis along y, $$c_E^2 = a^2 - b^2 = 169 - 144 = 25$$, giving $$c_E = 5$$ and foci at $$(0, \pm 5)$$.

For the hyperbola $$\frac{y^2}{\lambda^2} - \frac{x^2}{16} = 1$$, the focal distance satisfies $$c_H^2 = \lambda^2 + 16$$. Setting $$c_H = 5$$ yields $$25 = \lambda^2 + 16 \implies \lambda^2 = 9$$.

Its eccentricity is $$e = \frac{c_H}{\lambda} = \frac{5}{3}$$.

For the hyperbola $$\frac{y^2}{9} - \frac{x^2}{16} = 1$$, where $$a = 3$$ (transverse semi-axis) and $$b = 4$$, the length of the latus rectum is $$L = \frac{2b^2}{a} = \frac{2 \times 16}{3} = \frac{32}{3}$$.

Finally, computing $$24(e + L)$$ gives $$24\left(\frac{5}{3} + \frac{32}{3}\right) = 24 \times \frac{37}{3} = 8 \times 37 = 296$$.

The answer is Option C: $$296$$.

$$15(e^2 + 1) = 34e \implies 15e^2 - 34e + 15 = 0$$

$$\implies (5e - 3)(3e - 5) = 0$$

$$e = \frac{5}{3}$$ ($$e > 1$$)

$$b^2 = a^2\left(\frac{25}{9} - 1\right) \implies b^2 = \frac{16a^2}{9} \quad \text{--- (1)}$$

The hyperbola passes through the point $$(6, 4\sqrt{3})$$:

$$\frac{6^2}{a^2} - \frac{(4\sqrt{3})^2}{b^2} = 1 \implies \frac{36}{a^2} - \frac{48}{b^2} = 1$$

$$\frac{36}{a^2} - \frac{48}{\left(\frac{16a^2}{9}\right)} = 1$$

$$\frac{36}{a^2} - \frac{48 \times 9}{16a^2} = 1 \implies \frac{36}{a^2} - \frac{27}{a^2} = 1 \implies \frac{9}{a^2} = 1$$

$$\therefore a^2 = 9$$

$$b^2 = \frac{16(9)}{9} = 16$$

$$\frac{x^2}{b^2} - \frac{y^2}{2(a^2 + 1)} = 1$$

Comparing this with the standard form $$\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$$:

$$A^2 = b^2 = 16$$

$$B^2 = 2(a^2 + 1) = 2(9 + 1) = 20$$

The formula for the length of the latus rectum is: $$\text{Length} = \frac{2B^2}{A} = \frac{2(20)}{\sqrt{16}} = \frac{40}{4} = 10$$