Question

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.

• A. $(3\sqrt{2},2\sqrt{3})$
• B. $(2\sqrt{2},3\sqrt{3})$
• C. $(\sqrt{3},\sqrt{2})$
• D. $(-\sqrt{2},-\sqrt{3})$

Answer 1

Answer: (B)

$(2\sqrt{2},3\sqrt{3})$

Let equation of hypaerbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$(ae,0)=(\pm 2,0)$

Point P lies on hyperbola

So, $2b^2-3a^2=a^2{b}^2$ .........1

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

$b^2=4-a^2$ ....2

From (1) and (2)

$a^2=8,a^2=1$$,a^2\ne 8$

So, $b^2=4-1=3$

Hyperbola is $\frac{x^2}{1}-\frac{y^2}{3}=1$

$\frac{dy}{dx}=\frac{3x}{y}$

$\Rightarrow \frac{dy}{dx}=\sqrt{6}$ at P

Equation of tangent at P is $y-\sqrt{3}=\sqrt{6}(x-\sqrt{2})$

Option B $(2\sqrt{2},3\sqrt{3})$

$y-\sqrt{3}=\sqrt{6}(x-\sqrt{2})$

$3\sqrt{3}-\sqrt{3}=\sqrt{6}(2\sqrt{2}-\sqrt{2})$

$2\sqrt{3}=2\sqrt{3}$

Hence option B satisfies the equation.