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First of all, sorry for my poor English! Can you please help me? I'm trying to prove that, given a point P at an ellipse.

Please help me prove that the angles are equal.

Thanks! it's supposed to be an ellipse lol

Amzoti
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Mariana Ciotta
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2 Answers2

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It is not true for any point $P$. See the image:

enter image description here

Sigur
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  • So it must be at the vertice of the ellipse? – Mariana Ciotta May 15 '13 at 01:24
  • Note that $PO$ must be the bisectrix and the median of the triangle $F_1PF_2$. – Sigur May 15 '13 at 01:27
  • Oh, I got it! Thanks – Mariana Ciotta May 15 '13 at 01:30
  • I thought @MarianaCiotta was after the Reflection Property of ellipses. (A light ray emanating from one focus will bounce off the ellipse and head for the other focus.) Note the right angle marker in her diagram: this suggests (to me) that she's interested in the angle-dividing behavior of the perpendicular to the tangent line at $P$, not the behavior of line $OP$. In that case, it is true for any $P$: the perpendicular to the tangent always bisects the angle. – Blue May 15 '13 at 03:00
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You haven't stated the problem. You want to prove that a light ray from one focus reflects off the ellipse back to the other focus. You don't tell us what you know. Personally, I would represent the ellipse as a level set of the function $$f(\mathbf x) = \|\mathbf x-F_1\| + \|\mathbf x-F_2\|$$ and use the fact that $\nabla f$ is the normal vector.

You could also write down the equation $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$ and use implicit differentiation and lots of slopes.

Ted Shifrin
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