> Narasimham, > I used the definition of the ellipse F1 P + F2 P = 2a > just as you did to show that F1, P, F2 and Q are > vertices of a parallelogram, and rotational symmetry > to show that the tangents at P and Q are parallel. > I then invoked the known theorem that F1 P and F2 P > make equal angles with the tangent, and hence the > tangent bisects the external angles formed by the > focal radii at the point of contact. > I wanted to evaluate the angle in terms of cartesian > parameters, so I used the slope of the tangent, from > its equation, and the slopes of the focal radii to > calculate tan(theta). Using the facts that P lies on > the ellipse, and that a^2-b^2=c^2 I solved to give > tan(theta)= b^2/(c.y1) > Not very elegant, but effective. > Regards, Peter Scales.
Thanks it is clear now, your result is elegant indeed. At end of semi latus-rectum cot(theta)= eccentricity.