A hyperbola passes through the point $$P(\sqrt{2}, \sqrt{3})$$ and has foci at $$( \pm 2, 0)$$. Then the tangent to this hyperbola at $$P$$ also passes through the point

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.