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

Answer: 216

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