$Let P (3, 3) be a point on the hyperbola, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 . If the normal to it at P intersects the x -axis$
$at (9, 0) and e is its eccentricity, then the ordered pair (a^{2}, e^{2}) is equal to :$

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$(9, 3)$

$(\frac{3}{2}, 2)$

$(\frac{9}{2}, 2)$

$(\frac{9}{2}, 3)$

answer is D.

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Detailed Solution

$\begin{array}{l} P (3, 3) lies on \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{1}{a^{2}} - \frac{1}{b^{2}} = 9 . . . . . . . . . . . . . . (1) \\ Normal at P (3, 3) : \frac{a^{2}}{3} x + \frac{b^{2}}{3} y = a^{2} + b^{2} \end{array}$

$\begin{array}{l} (9, 0) lies \Rightarrow 2 a^{2} = b^{2} . . . . . . . . . . . . . . . . (2) \\ From (1) & (2) a^{2} = \frac{9}{2}; b^{2} = 9 \end{array}$