
Re: Same intersection angles in ellipse
Posted:
Oct 27, 2013 11:29 AM


> 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
Hi Narasimham,
the product of normals d1*d2 = const, also holds for any two isogonal conjugate points P,Q with respect to an arbitrary triangle, where d1,d2 denote distances of P,Q from a triangle side.
Best regards, Avni

