Let P(6,3) be a point on the hyperbolax2b2−y2b2=1 If the normal at the point P intersects the x-axis at (9, 0), then the eccentricity of the hyperbola is

x2a2−y2b2=1 or  2xa2−2yb2dydx=0 or  dydx=xb2ya2 Therefore, the slope of normal at (6,3) is −a2/2b2. The equation of normal is(y−3)=−a22b2(x−6)It passes through the point (9, 0). Therefore,a22b2=1or  e=32