JEE Main 2017 — Hyperbola Question with Solution

From the standard equation of hyperbola

$\Rightarrow \pm ae=\pm 2$

As we know, $b^2=a^2\left(e^2-1\right)$.

$\Rightarrow b^2=4-a^2$

$\therefore$Equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{4-a^2}=1$

$\because$It passes through $\left(\sqrt{2},\sqrt{3} \right)$

$\Rightarrow \frac{2}{a^2}-\frac{3}{4-a^2}=1$

$\Rightarrow a^2=t$

$\Rightarrow 2\left(4-t\right)-3t-t\left(4-t\right)=0$

$\Rightarrow 8-2t-3t-4t+t^2=0$

$\Rightarrow t^2-8t-t+8=0$

$\Rightarrow t= 8, 1$

So, $a^2=8, 1$.

$\Rightarrow a=2\sqrt{2}, 1$

But for $a=2\sqrt{2}, b$ becomes imaginary. So this case is rejected.

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

The equation of the tangent at $P$ is $\sqrt{2}x-\frac{y}{\sqrt{3}}=1$. 

Now checking each points.