Let P(3, 3) be a point on the hyperbola, $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$. If the normal to it at P intersects the x-axis at (9,0) and e is its eccentricity, then the ordered pair (a² , e²) is equal to :

 Given  equation of hyperbola is $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

Since Equation of hyperbola passes through (3, 3) then

$\frac{1}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}=\frac{1}{9}$    ……(i)

Equation of normal at point (3, 3) is :

$\frac{\mathrm{x}-3}{\frac{1}{{\mathrm{a}}^2}\cdot 3}=\frac{\mathrm{y}-3}{-\frac{1}{{\mathrm{b}}^2}\cdot 3}$ 

Since It passes through (9,0) then

$$\begin{gathered} \frac{6}{\frac{1}{{\mathrm{a}}^2}}=\frac{-3}{-\frac{1}{{\mathrm{b}}^2}} \\ \therefore \frac{1}{{\mathrm{b}}^2}=\frac{1}{2{\mathrm{a}}^2} \end{gathered}$$  ……(ii)

From equations (i) and (ii),

${\mathrm{a}}^2=\frac{9}{2},{\mathrm{b}}^2=9$

$\because$ Eccentricity = e, then ${\mathrm{e}}^2=1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=3$

$\therefore \left({\mathrm{a}}^2,{\mathrm{e}}^2\right)=\left(\frac{9}{2},3\right)$