Question

The vertices of a hyperbola $H$ are $(\pm 6,0)$ and its eccentricity is $\frac{\sqrt{5}}{2}$. Let $N$ be the normal to $H$ at a point in the first quadrant and parallel to the line $\sqrt{2} x+y=2 \sqrt{2}$. If $d$ is the length of the line segment of $N$ between $H$ and the $y$-axis then $d^2$ is equal to

Answer 1

$H : \frac{ x ^2}{36}-\frac{y^2}{9}=1$
equation of normal is $6 x \cos \theta+3 y \cot \theta=45$
$\text { slope }=-2 \sin \theta=-\sqrt{2} $
$ \Rightarrow \theta=\frac{\pi}{4}$
Equation of normal is $\sqrt{2} x + y =15$
$P :( a \sec \theta, b \tan \theta)$
$\Rightarrow P (6 \sqrt{2}, 3)$ and $K (0,15)$
$d^2=216$