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Topic: Same intersection angles in ellipse
Replies: 6   Last Post: Oct 29, 2013 3:11 PM

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Peter Scales

Posts: 192
From: Australia
Registered: 4/3/05
Re: Same intersection angles in ellipse
Posted: Oct 29, 2013 10:38 AM
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> > > 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:
> ? 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! )
> 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

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.

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