The vertices of a hyperbola H are ± 6 , 0 and its eccentricity is 5 2 . Let N be the normal to H at a…

Given equation of hyperbola is

$H:\frac{x^2}{36}-\frac{y^2}{9}=1$

Any point on the hyperbola is $P\left(6\sec \theta ,3\tan \theta \right)$.

Equation of normal is

$6x\cos \theta +3y\cot \theta =45$

Also, this normal is parallel to the line $\sqrt{2}x+y=2\sqrt{2}$, therefore slope of both lines must be equal , hence

$-2\sin \theta =-\sqrt{2}$

$\Rightarrow \sin \theta =\frac{1}{\sqrt{2}}$

$\Rightarrow \theta =\frac{\pi }{4}$

Equation of normal is

$\sqrt{2}x+y=15$

And, this normal intersects $y-$axis at $K$, so

$P\equiv \left(6\sqrt{2},3\right)$ and $K\equiv \left(0,15\right)$.

So,

$P{K}^2=d^2={\left(6\sqrt{2}-0\right)}^2+{\left(3-15\right)}^2$

$\Rightarrow d^2=216$