Let P(6, 3) be a point on the hyperbola (\frac{x^2}{a^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

Equation of normal to the hyperbola is (\frac{x - x_1}{x_1/a^2} = \frac{y - y_1}{-y_1/b^2}) So, equation of normal to hyperbola at (6,3) is (\frac{x - 6}{6/a^2} = \frac{y - 3}{-3/b^2}) Since, it intersects x-axis at (9,0) So, (\frac{9 - 6}{6/a^2} = \frac{-3}{-3/b^2} \implies a^2 = 2b^2) Eccentricity of hyperbola = (\sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{b^2}{2b^2}} = \sqrt{\frac{3}{2}})