The vertices of a hyperbola $H$ are $(±6, 0)$ 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

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answer is 216.

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Detailed Solution

$\begin{array}{l} a = 6, e = \frac{\sqrt{5}}{2} \\ ∴ \frac{5}{4} = 1 + \frac{b^{2}}{36} \Rightarrow b^{2} = 36 \times \frac{1}{4} = 9 \\ ∴ H : \frac{x^{2}}{36} - \frac{y^{2}}{9} = 1 \end{array}$

$\begin{array}{l} \begin{array}{l} P (6 \sec θ, 3 \tan θ); \\ Slope of tangent at P = \frac{6 \sec θ}{4 \times 3 \tan θ} \\ \frac{1}{2 \sin θ} \times - \sqrt{2} = - 1 \Rightarrow \sin θ = \frac{1}{\sqrt{2}} \Rightarrow θ = 45^{0} \\ P \equiv (6 \sqrt{2}, 3) \end{array} \\ \begin{array}{l} Q \equiv (0, 15) \\ ∴ {PQ}^{2} = 72 + 144 = 216 \end{array} \end{array}$