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

The correct answer is 216.

H:36x2​−9y2​=1
equation of normal is 6xcosθ+3ycotθ=45
slope =−2sinθ=−2​
⇒θ=4π​
Equation of normal is 2​x+y=15
P:(asecθ,btanθ)
⇒P(62​,3) and K(0,15)
d2=216