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

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Narasimham

Posts: 350
Registered: 9/16/06
Re: Same intersection angles in ellipse
Posted: Oct 26, 2013 9:46 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:
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



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