Let P(6,3) be a point on the hyperbola$\frac{x^2}{b^2}-\frac{y^2}{b^2}=1$ If the normal at the point P intersects the x-axis at (9, 0), then the eccentricity of the hyperbola is

$\begin{array}{l}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\\ \text{ or }\frac{2\mathrm{x}}{{\mathrm{a}}^2}-\frac{2\mathrm{y}}{{\mathrm{b}}^2}\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ \text{ or }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{xb}}^2}{{\mathrm{ya}}^2}\end{array}$

$\text{ Therefore, the slope of normal at }(6,3)\text{ is }-a^2/2b^2\text{. }$

The equation of normal is

$(y-3)=\frac{-a^2}{2b^2}(x-6)$

It passes through the point (9, 0). Therefore,
$\begin{array}{r}\frac{{\mathrm{a}}^2}{2{\mathrm{b}}^2}=1\\ \mathrm{or} \mathrm{e}=\sqrt{\frac{3}{2}}\end{array}$