> > Sum of of two focal rays lengths (from one focus) > > of an ellipse (semi major, minor axes a,b) is given as > > 2a. Show they make equal angles to the ellipse.. > > and find it.
> If I haven't misunderstood the question the two focal > rays from F1 meet the ellipse at diametrically > opposite points P1 and P2, from the definition of an > ellipse. > > The tangents at P1 and P2 are parallel, and bisect > the external angle between the focal rays. The figure > has rotational symmetry at Pi radians, so the angles > at P1(x1,y1) and P2(-x1,-y1) are equal to theta, > where > > tan(theta) = b^2/(c.y1) > > Regards, Peter Scales.
The result is correct, but unfortunately am readily unable to find which ellipse property you are referring to. May be property 2 in: http://www3.ul.ie/~rynnet/swconics/EP's.htm ? Please point to any diagram if possible.
F1 P+ F2 P = 2 a F1 P+ F1 Q = 2 a, given. So F1 Q = F2 P Likewise, F1 P = F2 Q F1, P, F2 and Q are vertices of a parallelogram and it is required to find inclination si of F1 P to tangent at P. Product of normals F1 P sin(si)* (2 a- F1 P) sin(si) = b^2 allows finding si for any F1 P.