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 $\left(a^{2}, e^{2}\right)$ is equal to :
Correct Option: 1
$\because$ The equation of hyperbola is
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$\because$ Equation of hyperbola passes through $(3,3)$
$\frac{1}{a^{2}}-\frac{1}{b^{2}}=\frac{1}{9}$ ...(i)
Equation of normal at point $(3,3)$ is :
$\frac{x-3}{\frac{1}{a^{2}} \cdot 3}=\frac{y-3}{-\frac{1}{b^{2}} \cdot 3}$
$\because$ It passes through $(9,0)$
$\frac{6}{\frac{1}{a^{2}}}=\frac{-3}{-\frac{1}{b^{2}}}$
$\therefore \frac{1}{b^{2}}=\frac{1}{2 a^{2}}$ ...(ii)
From equations (i) and (ii),
$a^{2}=\frac{9}{2}, b^{2}=9$
$\because$ Eccentricity $=e$, then $e^{2}=1+\frac{b^{2}}{a^{2}}=3$
$\therefore\left(a^{2}, e^{2}\right)=\left(\frac{9}{2}, 3\right)$