Let P(3, 3) be a point on the hyperbola, x ^ 2 a ^ 2 - y ^ 2 b ^ 2...

Given hyperbola is $\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1$.

$P\left(3, 3\right)$ lies on hyperbola then $\dfrac{1}{{a}^{2}} - \dfrac{1}{{b}^{2}} = \dfrac{1}{9} . . . \left(1\right)$

Normal at \left(3, 3\right)$is$\dfrac{{a}^{2}x}{3} + \dfrac{{b}^{2}y}{3} = {a}^{2} + {b}^{2}$</p><p>Normal passes through \left(9, 0\right)$

$3{a}^{2} = {a}^{2} + {b}^{2} ⇒ 2{a}^{2} = {b}^{2}$

Then using $( 1 )$, $\dfrac{1}{{a}^{2}} - \dfrac{1}{2{a}^{2}} = \dfrac{1}{9}$

$2{a}^{2} = 9 ⇒ {a}^{2} = \dfrac{9}{2}$ and ${b}^{2} = 9$

${e}^{2} = 1 + \dfrac{{b}^{2}}{{a}^{2}} = 1 + 2 = 3$.