> > > 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. > > Hi Peter, > > 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. > > I have used another property, viz., the product of > normals. ( It may be the same, but seen some other > way! ) > http://i42.tinypic.com/2afagbk.jpg > > 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. > > Best Regards, > Narasimham
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 canculate 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.