Ellipse Property Proof: (1 Viewer)

[ashokkumar]

Can someone show me the proof for the following (a diagram would also be helpful).

"A tangent at a point P on an ellipse is equally inclined to the focal chords through P."

thanks in advance.

 

[ashokkumar]

I tried, but i can't get it.

 

[Trebla]

I think the Cambridge textbook or some textbook (can't remember) has a proof of that in the examples because it is a syllabus dot point proof. It's actually quite long.

 

[seanieg89]

The Cambridge book has a proof based on coordinate geometry and several leadup questions. There is a shorter proof based on the (equal sum of focal lengths) property of the ellipse, along with the triangle inequality. I wrote it up formally for a student once before but this is the outline:

1. Reflect one focus S about the tangent, call this new point T.

2. Prove SQM and TQM are congruent triangles, where M is the midpoint of S,T (and hence lies on the tangent), and Q is an arbitrary point on the tangent.

3. The tangent lies external to the ellipse hence for all points Q on it, SQ+S'Q>2a, with equality if and only if Q=P.

4. Hence S'Q+TQ minimised at Q=P. But by the triangle inequality, this implies that P lies on the straight line S'T.

5. Hence S',P,T collinear. This gives us enough geometric information to angle chase.

This argument can be rigorised.

 

[nat_doc]

Its in the terry lee