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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 29, 2013 3:11 PM
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> 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
> calculate 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.


Thanks it is clear now, your result is elegant indeed. At end of semi latus-rectum cot(theta)= eccentricity.

Regards
Narasimham



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