Prove the sum of the distances from any point of an ellipse to its foci is constant.

by samer abdullah   Last Updated December 07, 2017 10:20 AM

A conic symmetric to the origin satisfies the equation $\Vert X-F \Vert =\vert eX\cdot N -\alpha \vert$ ,where $ \alpha = ed+F\cdot N $.(d is the distance from F to the directix). Use this relation to prove that $\Vert X-F \Vert + \Vert X+F \Vert =2\alpha $ if the conic is an ellipse.

Proof, Since the conic is symmetric the first equation is valid for -X then, $\Vert -X-F \Vert=\vert -eX\cdot N-\alpha \vert $ so $\Vert X+F \Vert =\vert eX\cdot N+\alpha \vert$ and $\Vert X-F\Vert +\Vert X+F\Vert =\vert eX\cdot N -\alpha \vert +\vert eX\cdot N+\alpha \vert $ It seems it suffices to show $\vert X\cdot N \vert \le \alpha $ then $ e\vert X\cdot N \vert \le \alpha $ because $e\lt 1$. From the sketch of an ellipse $\vert X\cdot N \vert \le \alpha $ seems true but I cannot prove it . I tried to show that if $\vert eX\cdot N\vert \gt \alpha $ then $ \vert X\cdot N \vert\gt d+F\cdot N$ but that would mean that X lies in the right side of the directix wich is valid only for a hyperbola (left side of the left directix if XN is negative).I wonder if the proof is complete. thank you in advance!

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