JEE Locus Questions

We need to find the eccentricity of the ellipse formed by the locus of the midpoint of the line segment joining $$(4, 3)$$ and points on the ellipse $$x^2 + 2y^2 = 4$$.

The ellipse $$x^2 + 2y^2 = 4$$ can be written as $$\frac{x^2}{4} + \frac{y^2}{2} = 1$$ so a general point on it is $$(2\cos\theta, \sqrt{2}\sin\theta)$$.

If we denote the midpoint of the segment from $$(4,3)$$ to $$(2\cos\theta,\sqrt{2}\sin\theta)$$ by $$(h,k)$$, then

$$h = \frac{4 + 2\cos\theta}{2} = 2 + \cos\theta$$

$$k = \frac{3 + \sqrt{2}\sin\theta}{2}$$

From these expressions we have $$\cos\theta = h - 2$$ and $$\sin\theta = \frac{2k - 3}{\sqrt{2}}$$.

Using the identity $$\cos^2\theta + \sin^2\theta = 1$$ yields

$$(h - 2)^2 + \frac{(2k - 3)^2}{2} = 1$$.

This can be rewritten in standard form as

$$\frac{(h - 2)^2}{1} + \frac{\left(k - \frac{3}{2}\right)^2}{\frac{1}{2}} = 1$$,

so that $$a^2 = 1$$ and $$b^2 = \frac{1}{2}$$ with $$a^2 > b^2$$.

The eccentricity is then

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

The answer is Option C: $$\dfrac{1}{\sqrt{2}}$$.

Let the fixed end of the line segment be the point $$(-3,\,-5)$$.

We take an arbitrary point on the ellipse $$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=1$$ and denote its coordinates by $$(x_{1},\,y_{1})$$.

The mid-point $$(h,\,k)$$ of the segment joining $$(-3,\,-5)$$ and $$(x_{1},\,y_{1})$$ is obtained from the section formula. We have

$$h=\dfrac{-3+x_{1}}{2},\qquad k=\dfrac{-5+y_{1}}{2}.$$

Now we express $$(x_{1},\,y_{1})$$ in terms of $$(h,\,k)$$ by simply reversing the above relations:

$$x_{1}=2h+3,\qquad y_{1}=2k+5.$$

Because $$(x_{1},\,y_{1})$$ lies on the ellipse, it must satisfy the given equation. So we substitute these expressions:

$$\frac{(2h+3)^{2}}{4}+\frac{(2k+5)^{2}}{9}=1.$$

We expand the squares first:

$$(2h+3)^{2}=4h^{2}+12h+9,$$

$$(2k+5)^{2}=4k^{2}+20k+25.$$

Substituting these back, we get

$$\frac{4h^{2}+12h+9}{4}+\frac{4k^{2}+20k+25}{9}=1.$$

Now we divide term-wise:

$$\bigl(4h^{2}+12h+9\bigr)\!\left(\frac{1}{4}\right)=h^{2}+3h+\frac{9}{4},$$

$$\bigl(4k^{2}+20k+25\bigr)\!\left(\frac{1}{9}\right)=\frac{4}{9}k^{2}+\frac{20}{9}k+\frac{25}{9}.$$

Adding these two expressions and subtracting $$1$$ from both sides, we write

$$h^{2}+3h+\frac{9}{4}+\frac{4}{9}k^{2}+\frac{20}{9}k+\frac{25}{9}-1=0.$$

Next we gather the constant terms. To combine the fractions, we use a common denominator $$36$$:

$$\frac{9}{4}=\frac{81}{36},\qquad\frac{25}{9}=\frac{100}{36},\qquad 1=\frac{36}{36}.$$

So

$$\frac{81}{36}+\frac{100}{36}-\frac{36}{36}=\frac{145}{36}.$$

Hence the equation becomes

$$h^{2}+3h+\frac{4}{9}k^{2}+\frac{20}{9}k+\frac{145}{36}=0.$$

To clear all fractions, we multiply every term by $$36$$:

$$36h^{2}+108h+16k^{2}+80k+145=0.$$

Finally, we replace the temporary symbols $$h$$ and $$k$$ by the usual coordinate variables $$x$$ and $$y$$ for the locus:

$$36x^{2}+16y^{2}+108x+80y+145=0.$$

This equation matches option B.

Hence, the correct answer is Option B.

We start from the given hyperbola

$$16x^{2}-9y^{2}+32x+36y-164=0.$$

First we bring it to its standard form by completing the squares. We collect the $$x$$-terms together and the $$y$$-terms together:

$$16x^{2}+32x-\,9y^{2}+36y-164=0.$$

Taking $$16$$ common from the $$x$$-terms and $$-9$$ common from the $$y$$-terms, we get

$$16\bigl(x^{2}+2x\bigr)-9\bigl(y^{2}-4y\bigr)-164=0.$$

We now complete the square in each bracket.

For $$x^{2}+2x$$ we use $$x^{2}+2x=\bigl(x+1\bigr)^{2}-1.$$

For $$y^{2}-4y$$ we use $$y^{2}-4y=\bigl(y-2\bigr)^{2}-4.$$

Substituting these we have

$$16\Bigl[\bigl(x+1\bigr)^{2}-1\Bigr] \;-\;9\Bigl[\bigl(y-2\bigr)^{2}-4\Bigr]\;-\;164=0.$$

Expanding the constants:

$$16\bigl(x+1\bigr)^{2}-16-9\bigl(y-2\bigr)^{2}+36-164=0.$$

The constant terms combine to $$-16+36-164=-144$$, hence

$$16\bigl(x+1\bigr)^{2}-9\bigl(y-2\bigr)^{2}-144=0.$$

We divide by $$144$$ so that the right-hand side becomes $$1$$:

$$\frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1.$$

Thus the hyperbola is centered at $$C(-1,\,2)$$ with

$$a^{2}=9,\; a=3,\qquad b^{2}=16,\; b=4.$$

Because the $$x$$-term is positive, the transverse axis is along the $$x$$-direction. For a hyperbola of the form $$\dfrac{(x-h)^{2}}{a^{2}}-\dfrac{(y-k)^{2}}{b^{2}}=1$$ the distance of each focus from the centre is given by

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

So here

$$c^{2}=9+16=25 \;\Longrightarrow\; c=5.$$

Hence the two foci are

$$F_{1}\bigl(-1-c,\,2\bigr)=(-6,\,2),\qquad F_{2}\bigl(-1+c,\,2\bigr)=(4,\,2).$$

Let $$P(x,\,y)$$ be any point on the hyperbola. The centroid $$G(h,\,k)$$ of the triangle with vertices $$P$$, $$F_{1}$$ and $$F_{2}$$ is obtained by the formula

$$h=\frac{x_{P}+x_{F_{1}}+x_{F_{2}}}{3}, \qquad k=\frac{y_{P}+y_{F_{1}}+y_{F_{2}}}{3}.$$

Substituting the coordinates, we obtain

$$h=\frac{x-6+4}{3}=\frac{x-2}{3},\qquad k=\frac{y+2+2}{3}=\frac{y+4}{3}.$$

This immediately gives $$x$$ and $$y$$ in terms of $$h$$ and $$k$$:

$$x=3h+2,\qquad y=3k-4.$$

Because $$P(x,y)$$ lies on the hyperbola, it must satisfy

$$\frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1.$$

We now substitute $$x=3h+2$$ and $$y=3k-4$$ into this equation.

The $$x$$-part becomes

$$x+1=3h+2+1=3h+3=3(h+1).$$

Hence

$$\frac{(x+1)^{2}}{9}=\frac{\bigl(3(h+1)\bigr)^{2}}{9} =\frac{9(h+1)^{2}}{9}= (h+1)^{2}.$$

Similarly, the $$y$$-part is

$$y-2=3k-4-2=3k-6=3(k-2),$$

so that

$$\frac{(y-2)^{2}}{16}=\frac{\bigl(3(k-2)\bigr)^{2}}{16} =\frac{9(k-2)^{2}}{16}.$$

Putting these back into the hyperbola equation we get

$$(h+1)^{2}-\frac{9(k-2)^{2}}{16}=1.$$

To remove the denominator, we multiply by $$16$$:

$$16(h+1)^{2}-9(k-2)^{2}=16.$$

Next we expand each square. First,

$$16(h+1)^{2}=16(h^{2}+2h+1)=16h^{2}+32h+16,$$

and

$$9(k-2)^{2}=9(k^{2}-4k+4)=9k^{2}-36k+36.$$

Substituting these expansions, the equation becomes

$$\bigl(16h^{2}+32h+16\bigr)-\bigl(9k^{2}-36k+36\bigr)=16.$$

We now open the bracket with the minus sign:

$$16h^{2}+32h+16-9k^{2}+36k-36=16.$$

Finally, we bring the $$16$$ on the right over to the left so that the whole expression equals zero:

$$16h^{2}+32h-9k^{2}+36k-36=0.$$

The centroid coordinates are merely dummy variables, so we rename them $$(x,y)$$ to state the locus in the usual form:

$$16x^{2}-9y^{2}+32x+36y-36=0.$$

This equation matches Option A. Hence, the correct answer is Option A.

Let the mid-point of a variable chord of the hyperbola $$x^{2}-y^{2}=4$$ be $$(h,k)$$. For any conic whose equation is written as $$S(x,y)=0,$$ the equation of the chord whose mid-point is $$(h,k)$$ is obtained from the formula $$T=S(h,k).$$

We first write the hyperbola in the form $$S(x,y)=0$$:

$$S(x,y)=x^{2}-y^{2}-4=0.$$

Replacing $$x^{2}$$ by $$x\,h$$ and $$y^{2}$$ by $$y\,k$$ we get

$$T=xh-\;y k-4.$$

Next we evaluate $$S(h,k)$$:

$$S(h,k)=h^{2}-k^{2}-4.$$

The required chord therefore satisfies

$$T=S(h,k)\;\;\Longrightarrow\;\;xh-yk-4=h^{2}-k^{2}-4.$$

Simplifying, the chord is

$$h\,x-k\,y=h^{2}-k^{2}.\qquad(1)$$

This straight line is given to be tangent to the parabola $$y^{2}=8x.$$ For the standard parabola $$y^{2}=4ax$$ (here $$4a=8\;\Rightarrow\;a=2$$) the slope form of a tangent is

$$y=mx+\frac{a}{m}=mx+\frac{2}{m}.$$(2)

We now bring equation (1) to the same form so that we may compare the coefficients. Solving (1) for $$y$$ we have

$$k\,y=h\,x-(h^{2}-k^{2})$$

$$\Longrightarrow\;y=\frac{h}{k}\,x-\frac{h^{2}-k^{2}}{k}.$$(3)

From (3) the slope is

$$m=\frac{h}{k},$$

and the $$x$$-intercept term is

$$c=-\frac{h^{2}-k^{2}}{k}.$$

Because (3) must coincide with the general tangent (2), both the slopes and the constant terms must match. We already have the equality of slopes through $$m=\dfrac{h}{k}.$$ Equating the constant terms,

$$-\frac{h^{2}-k^{2}}{k}=\frac{2}{m}=\frac{2k}{h}.$$

Clearing denominators step by step:

$$-\bigl(h^{2}-k^{2}\bigr)=\frac{2k^{2}}{h}$$

$$\Longrightarrow\;-h\bigl(h^{2}-k^{2}\bigr)=2k^{2}$$

$$\Longrightarrow\;-h^{3}+h\,k^{2}=2k^{2}$$

$$\Longrightarrow\;h^{3}-h\,k^{2}+2k^{2}=0.$$

Collecting $$k^{2}$$ as a common factor in the last two terms gives

$$h^{3}=k^{2}(h-2).$$

Finally, replacing the fixed parameters $$h$$ and $$k$$ by the general coordinates $$x$$ and $$y$$ of the sought locus, we obtain

$$y^{2}(x-2)=x^{3}.$$

This is exactly the equation listed in Option A.

Hence, the correct answer is Option A.

Let the midpoint of a chord of the circle $$x^2 + y^2 = 25$$ be $$(h, k)$$. The equation of the chord with this midpoint is $$hx + ky = h^2 + k^2$$.

Rewriting this as a line in slope-intercept form: $$y = -\frac{h}{k}x + \frac{h^2 + k^2}{k}$$, where the slope is $$m = -\frac{h}{k}$$ and the y-intercept is $$c = \frac{h^2 + k^2}{k}$$.

For this line to be tangent to the hyperbola $$\frac{x^2}{9} - \frac{y^2}{16} = 1$$, the tangency condition requires $$c^2 = 9m^2 - 16$$. Substituting the values of $$m$$ and $$c$$:

$$\frac{(h^2 + k^2)^2}{k^2} = 9 \cdot \frac{h^2}{k^2} - 16$$

Multiplying both sides by $$k^2$$: $$(h^2 + k^2)^2 = 9h^2 - 16k^2$$.

Replacing $$(h, k)$$ with $$(x, y)$$, the required locus is $$(x^2 + y^2)^2 - 9x^2 + 16y^2 = 0$$.

We let the moving point be $$P(x,y)$$. The fixed points are $$O(0,0)$$ and $$A(0,1)$$. The condition given is that the perimeter of $$\triangle AOP$$ is $$4$$.

By the distance formula, the length of a segment whose end-points are $$(x_1,y_1)$$ and $$(x_2,y_2)$$ equals $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$. Using this formula one by one we obtain

$$OA=\sqrt{(0-0)^2+(1-0)^2}=1,$$

$$OP=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^{2}+y^{2}},$$

$$AP=\sqrt{(x-0)^2+(y-1)^2}=\sqrt{x^{2}+(y-1)^{2}}.$$

The perimeter condition therefore reads

$$OA+OP+AP=4 \;\;\Longrightarrow\;\; 1+\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+(y-1)^{2}}=4.$$

Subtracting $$1$$ from both sides gives

$$\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+(y-1)^{2}}=3.$$

We now remove the square-roots step by step. First we square once:

$$\bigl(\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+(y-1)^{2}}\bigr)^{2}=9.$$

Using $$(a+b)^{2}=a^{2}+2ab+b^{2},$$ we get

$$x^{2}+y^{2}+2\sqrt{(x^{2}+y^{2})\bigl(x^{2}+(y-1)^{2}\bigr)}+x^{2}+(y-1)^{2}=9.$$

Combining the like terms inside the radical first:

$$(y-1)^{2}=y^{2}-2y+1,$$

so the non-radical part becomes

$$2x^{2}+y^{2}+y^{2}-2y+1 = 2x^{2}+2y^{2}-2y+1.$$

Hence the equation is

$$2x^{2}+2y^{2}-2y+1+2\sqrt{(x^{2}+y^{2})\bigl(x^{2}+y^{2}-2y+1\bigr)}=9.$$

Isolating the radical term:

$$2\sqrt{(x^{2}+y^{2})\bigl(x^{2}+y^{2}-2y+1\bigr)}=9-\bigl(2x^{2}+2y^{2}-2y+1\bigr).$$

Simplifying the right side:

$$9-(2x^{2}+2y^{2}-2y+1)=8-2x^{2}-2y^{2}+2y.$$

Dividing by $$2$$ gives

$$\sqrt{(x^{2}+y^{2})\bigl(x^{2}+y^{2}-2y+1\bigr)}=4-x^{2}-y^{2}+y.$$

We square a second time to eliminate the remaining root:

$$(x^{2}+y^{2})\bigl(x^{2}+y^{2}-2y+1\bigr)=\bigl(4-x^{2}-y^{2}+y\bigr)^{2}.$$

For convenience set $$A=x^{2}+y^{2}$$. Then the left side becomes

$$A(A-2y+1)=A^{2}-2yA+A.$$

Expanding back in terms of $$x$$ and $$y$$ we obtain

$$x^{4}+2x^{2}y^{2}+y^{4}-2x^{2}y-2y^{3}+x^{2}+y^{2}.$$

Now we expand the right side. Write $$B=4-x^{2}-y^{2}+y$$ and use $$(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2(ab+ac+ad+bc+bd+cd).$$ With $$a=4,\; b=-x^{2},\; c=-y^{2},\; d=y$$ we get

$$\bigl(4-x^{2}-y^{2}+y\bigr)^{2}=16+x^{4}+y^{4}+y^{2}-8x^{2}-8y^{2}+8y+2x^{2}y^{2}-2x^{2}y-2y^{3}.$$

Equating the left and right expansions and cancelling the common terms $$x^{4},\; 2x^{2}y^{2},\; y^{4},\; -2x^{2}y,\; -2y^{3}$$ gives the simpler relation

$$-8x^{2}-7y^{2}+8y+16-x^{2}-y^{2}=0.$$

Combining like terms leads to

$$-9x^{2}-8y^{2}+8y+16=0.$$

Multiplying by $$-1$$ (which does not change the locus) puts it in the customary form

$$9x^{2}+8y^{2}-8y-16=0.$$

Finally, moving the constant term to the right side we have

$$9x^{2}+8y^{2}-8y=16.$$

This equation exactly matches the expression given in Option D:

$$9x^{2}+8y^{2}-8y=16.$$

Hence, the correct answer is Option D.

We begin with the hyperbola $$4x^2-9y^2 = 36$$ and choose an arbitrary point on it, say $$P_1(x_1 ,\, y_1)$$. Because this point lies on the curve we already have

$$4x_1^2-9y_1^2 = 36 \;. \quad -(1)$$

To write the normal at this point, we first need the slope of the tangent. Differentiating $$4x^2-9y^2 = 36$$ implicitly with respect to $$x$$ we get

$$8x - 18y\,\dfrac{dy}{dx}=0,$$

so the derivative (slope of the tangent) is

$$\dfrac{dy}{dx} = \dfrac{8x}{18y} = \dfrac{4x}{9y}.$$

At the chosen point $$P_1(x_1,y_1)$$ the tangent slope is therefore

$$m_t = \dfrac{4x_1}{9y_1}.$$

The slope of the normal is the negative reciprocal of the tangent slope; hence

$$m_n = -\,\dfrac{1}{m_t}= -\,\dfrac{9y_1}{4x_1}.$$

The equation of the normal line through $$P_1(x_1,y_1)$$ then becomes

$$y-y_1 = -\,\dfrac{9y_1}{4x_1}\,(x-x_1). \quad -(2)$$

Next we find where this normal meets the coordinate axes.

x-intercept A (set $$y = 0$$ in (2)):

$$0 - y_1 = -\,\dfrac{9y_1}{4x_1}\,(x_A - x_1).$$

Dividing by $$-y_1$$ (assuming $$y_1 \neq 0$$) and simplifying,

$$1 = \dfrac{9}{4x_1}\,(x_A - x_1) \;\;\Longrightarrow\;\; x_A - x_1 = \dfrac{4x_1}{9} \;\;\Longrightarrow\;\; x_A = x_1 + \dfrac{4x_1}{9} = \dfrac{13x_1}{9}.$$

y-intercept B (set $$x = 0$$ in (2)):

$$y_B - y_1 = -\,\dfrac{9y_1}{4x_1}\,(0 - x_1) = \dfrac{9y_1}{4}.$$

Hence

$$y_B = y_1 + \dfrac{9y_1}{4} = \dfrac{13y_1}{4}.$$

Thus the intercepts are

$$A\left(\dfrac{13x_1}{9},\,0\right),\qquad B\left(0,\,\dfrac{13y_1}{4}\right).$$

The problem now forms the parallelogram $$OABP$$ with the origin $$O(0,0)$$. In a parallelogram, the fourth vertex is obtained by the vector sum of the position vectors of the adjacent vertices, so

$$\vec{OP} = \vec{OA} + \vec{OB}.$$

Consequently, the coordinates of $$P$$ are

$$P\;(x,y) = \left(\dfrac{13x_1}{9},\;\dfrac{13y_1}{4}\right).$$

We now express $$x_1$$ and $$y_1$$ in terms of $$x$$ and $$y$$:

$$x_1 = \dfrac{9x}{13},\qquad y_1 = \dfrac{4y}{13}.$$

Substituting these into the point condition (1) gives the required locus:

$$ 4\left(\dfrac{9x}{13}\right)^2 - 9\left(\dfrac{4y}{13}\right)^2 = 36. $$

Carrying out the squares,

$$ 4 \cdot \dfrac{81x^2}{169} \;-\; 9 \cdot \dfrac{16y^2}{169} = 36 \;\;\Longrightarrow\;\; \dfrac{324x^2}{169} \;-\; \dfrac{144y^2}{169} = 36. $$

Multiplying by $$169$$ clears the denominators:

$$324x^2 - 144y^2 = 36 \times 169 = 6084.$$

Finally we divide every term by $$36$$ to simplify:

$$9x^2 - 4y^2 = 169.$$

This equation represents a hyperbola and matches Option 3.

Hence, the correct answer is Option 3.

We start from the given hyperbola $$4y^2 = x^2 + 1$$ and write the equation of any line in slope-intercept form as $$y = mx + c$$, where $$m$$ is the slope and $$c$$ is the $$y$$-intercept.

For this line to touch the hyperbola, it must satisfy the condition of tangency. We substitute $$y = mx + c$$ into the hyperbola:

$$4(mx + c)^2 = x^2 + 1.$$

Expanding the left side we obtain

$$4\bigl(m^2x^2 + 2mcx + c^2\bigr) = x^2 + 1,$$ so $$4m^2x^2 + 8mcx + 4c^2 = x^2 + 1.$$ Bringing all terms to one side,

$$(4m^2 - 1)x^2 + 8mcx + (4c^2 - 1) = 0.$$ This is a quadratic in $$x$$. For the line to be a tangent, its discriminant must be zero. The discriminant formula is $$\Delta = B^2 - 4AC$$ for the quadratic $$Ax^2 + Bx + C = 0$$. Here we have

$$A = 4m^2 - 1,\quad B = 8mc,\quad C = 4c^2 - 1.$$

Setting the discriminant to zero gives

$$B^2 - 4AC = 0,$$ so $$(8mc)^2 - 4(4m^2 - 1)(4c^2 - 1) = 0.$$ Dividing by 4 to simplify,

$$16m^2c^2 - (4m^2 - 1)(4c^2 - 1) = 0.$$

Now we expand the product: $$(4m^2 - 1)(4c^2 - 1) = 16m^2c^2 - 4m^2 - 4c^2 + 1,$$ and substitute back:

$$16m^2c^2 - \bigl[16m^2c^2 - 4m^2 - 4c^2 + 1\bigr] = 0.$$ The terms $$16m^2c^2$$ cancel, leaving

$$4m^2 + 4c^2 - 1 = 0,$$ or after dividing by 4,

$$m^2 + c^2 = \dfrac14.$$

Thus any tangent to the hyperbola can be written as $$y = mx + c$$ with $$m^2 + c^2 = \dfrac14.$$

Such a tangent meets the coordinate axes at the points $$A\bigl(x_A, 0\bigr) \quad\text{and}\quad B\bigl(0, c\bigr).$$ Indeed, at $$y = 0$$ we get $$x_A = -\dfrac{c}{m},$$ and at $$x = 0$$ we get $$y = c.$$

Let $$M(h,k)$$ be the midpoint of the segment $$AB$$. Midpoint formulas give

$$h = \dfrac{x_A + 0}{2} = \dfrac{-c/m}{2} = -\dfrac{c}{2m},$$ $$k = \dfrac{0 + c}{2} = \dfrac{c}{2}.$$

From the second relation we have $$c = 2k.$$ Substituting this into the first gives

$$h = -\dfrac{2k}{2m} = -\dfrac{k}{m},$$ so $$m = -\dfrac{k}{h}.$$

We now insert $$m = -\dfrac{k}{h}$$ and $$c = 2k$$ into the tangency condition $$m^2 + c^2 = \dfrac14$$:

$$\left(-\dfrac{k}{h}\right)^2 + (2k)^2 = \dfrac14.$$ Thus $$\dfrac{k^2}{h^2} + 4k^2 = \dfrac14.$$

Multiplying through by $$h^2$$ eliminates the denominator:

$$k^2 + 4k^2h^2 = \dfrac{h^2}{4}.$$

Re-arranging all terms to one side,

$$k^2 + 4k^2h^2 - \dfrac{h^2}{4} = 0.$$

Multiplying by 4 to clear the remaining fraction,

$$4k^2 + 16k^2h^2 - h^2 = 0.$$

Writing the terms in a more standard order gives

$$-h^2 + 4k^2 + 16h^2k^2 = 0,$$ and multiplying by $$-1$$ yields

$$h^2 - 4k^2 - 16h^2k^2 = 0.$$

Since $$(h,k)$$ represents the general midpoint, we now replace $$h$$ by $$x$$ and $$k$$ by $$y$$ to obtain the required locus:

$$x^2 - 4y^2 - 16x^2y^2 = 0.$$

Comparing with the options, this matches Option D.

Hence, the correct answer is Option D.

We start with the given ellipse

$$x^{2}+3y^{2}=6.$$

It is convenient to write it in the standard form

$$\frac{x^{2}}{6}+\frac{y^{2}}{2}=1,$$

so that we can recognise the semi-axes $$a^{2}=6$$ and $$b^{2}=2.$$

For any point $$\,(x_{1},y_{1})$$ lying on the ellipse, i.e.

$$x_{1}^{2}+3y_{1}^{2}=6,$$

the equation of the tangent at that point is obtained by the tangent in the form $$T=0$$ formula:

$$xx_{1}+3yy_{1}=6.$$

The centre of the ellipse is the origin $$O(0,0).$$ We are required to find the locus of the foot of the perpendicular drawn from this origin to an arbitrary tangent. Let the foot of that perpendicular be $$P(h,k).$$

Because $$OP$$ is perpendicular to the tangent, the vector $$\overrightarrow{OP}=(h,k)$$ must be parallel to the normal vector of the tangent. For the line

$$xx_{1}+3yy_{1}=6,$$

the normal vector is clearly $$\bigl(x_{1},\,3y_{1}\bigr).$$ Hence there is some scalar $$t$$ such that

$$h=t\,x_{1},\qquad k=t\,(3y_{1}).$$

Point $$P(h,k)$$ itself lies on the tangent, so we substitute $$x=h,\;y=k$$ into the tangent’s equation:

$$x_{1}h+3y_{1}k=6.$$ Substituting $$h=t\,x_{1}$$ and $$k=t\,3y_{1}$$ gives

$$x_{1}(t\,x_{1})+3y_{1}(t\,3y_{1})=t\bigl(x_{1}^{2}+9y_{1}^{2}\bigr)=6.$$

Therefore

$$t=\frac{6}{x_{1}^{2}+9y_{1}^{2}}.$$

Using this value of $$t$$ we can express $$h$$ and $$k$$ completely in terms of $$x_{1},y_{1}:$$

$$h=\frac{6x_{1}}{x_{1}^{2}+9y_{1}^{2}},\qquad k=\frac{18y_{1}}{x_{1}^{2}+9y_{1}^{2}}.$$

Our goal is to eliminate $$x_{1},y_{1}$$ to obtain a relation between $$h$$ and $$k$$ only. To that end we let

$$D=x_{1}^{2}+9y_{1}^{2},$$

so that the above formulae become

$$h=\frac{6x_{1}}{D},\qquad k=\frac{18y_{1}}{D}.$$

From these we solve for $$x_{1}$$ and $$y_{1}$$:

$$x_{1}=\frac{hD}{6},\qquad y_{1}=\frac{kD}{18}.$$

Since $$\,(x_{1},y_{1})$$ lies on the ellipse, it satisfies

$$x_{1}^{2}+3y_{1}^{2}=6.$$

Substituting the expressions for $$x_{1}$$ and $$y_{1}$$ gives

$$\left(\frac{hD}{6}\right)^{2}+3\left(\frac{kD}{18}\right)^{2}=6.$$

Expanding, we obtain

$$\frac{h^{2}D^{2}}{36}+\frac{k^{2}D^{2}}{108}=6.$$

Factorising $$\dfrac{1}{108}$$ out of the brackets,

$$\frac{D^{2}}{108}\bigl(3h^{2}+k^{2}\bigr)=6,$$ so

$$D^{2}=\frac{648}{3h^{2}+k^{2}}.$$

On the other hand, by definition

$$D=x_{1}^{2}+9y_{1}^{2}=\frac{h^{2}D^{2}}{36}+\frac{k^{2}D^{2}}{36} = \frac{D^{2}(h^{2}+k^{2})}{36}.$$

Dividing both sides by $$D\ (\neq 0)$$ gives

$$1=\frac{D(h^{2}+k^{2})}{36},\qquad\text{so}\qquad D=\frac{36}{h^{2}+k^{2}}.$$

Squaring this last relation,

$$D^{2}=\frac{1296}{(h^{2}+k^{2})^{2}}.$$

We now have two separate expressions for $$D^{2}$$, so we equate them:

$$\frac{1296}{(h^{2}+k^{2})^{2}}=\frac{648}{3h^{2}+k^{2}}.$$

Cross-multiplying yields

$$1296\bigl(3h^{2}+k^{2}\bigr)=648\,(h^{2}+k^{2})^{2}.$$

Dividing by $$648$$ simplifies the equation to

$$2\bigl(3h^{2}+k^{2}\bigr)=(h^{2}+k^{2})^{2}.$$

That is,

$$(h^{2}+k^{2})^{2}=6h^{2}+2k^{2}.$$

Replacing $$h$$ by $$x$$ and $$k$$ by $$y$$ (for the locus in the usual $$xy$$-plane) we finally obtain

$$(x^{2}+y^{2})^{2}=6x^{2}+2y^{2}.$$

This exactly matches the equation given in Option A.

Hence, the correct answer is Option A.

The given family of parabolas is
$$y=\frac{a^{3}x^{2}}{3}+\frac{a^{2}x}{2}-2a \qquad (a\in\mathbb{R})$$

For any parabola written as $$y=Ax^{2}+Bx+C$$ the vertex $$V(x_{v},y_{v})$$ is obtained from

$$x_{v}=-\frac{B}{2A}, \qquad y_{v}=C-\frac{B^{2}}{4A}$$

Here
$$A=\frac{a^{3}}{3},\qquad B=\frac{a^{2}}{2},\qquad C=-2a$$

Step 1: x-coordinate of the vertex
$$x=-\frac{B}{2A}=-\frac{\dfrac{a^{2}}{2}}{2\left(\dfrac{a^{3}}{3}\right)}=-\frac{3}{4a}$$

Step 2: y-coordinate of the vertex
$$$ \begin{aligned} y&=C-\frac{B^{2}}{4A}\\ &=-2a-\frac{\left(\dfrac{a^{2}}{2}\right)^{2}}{4\left(\dfrac{a^{3}}{3}\right)}\\ &=-2a-\frac{a^{4}/4}{4a^{3}/3}\\ &=-2a-\frac{3a}{16}\\ &=-\frac{35a}{16} \end{aligned} $$$

Thus every vertex has coordinates
$$\bigl(x,\;y\bigr)=\left(-\frac{3}{4a},\;-\frac{35a}{16}\right).$$

Step 3: Eliminate the parameter $$a$$

From $$x=-\dfrac{3}{4a}\; \Rightarrow\; a=-\dfrac{3}{4x}$$

Substitute this value in the expression for $$y$$:
$$y=-\frac{35}{16}\left(-\frac{3}{4x}\right)=\frac{105}{64}\,\frac{1}{x}$$

Re-arranging, we obtain the locus

$$xy=\frac{105}{64}.$$

Hence the locus of the vertices is $$xy=\dfrac{105}{64}$$.

Option A which is: $$xy = \dfrac{105}{64}$$