I came across the following property of the ellipse:

The distance from a focus of an ellipse to any point on the ellipse is equal to $a(1-e \cos\theta)$. Where the $a$ is the length of semi-major axis and $\theta$ is the eccentric angle of the point.

I can prove this with coordinate geometry but I want a pure geometric proof of it. Please help.


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Here $O$ is the centre, and $F$ is a focus. It is known that $OF=ae$,$PN=b\sin\theta$ and $OQ=a$. Apply Pythagoras Theorem in the $\triangle PNF$: $$PF^2=PN^2+NF^2=PN^2+(OF-ON)^2=PN^2+(OF-OQ\cos\theta)^2$$ $$=b^2\sin^2\theta+(ae-a\cos\theta)^2=a^2\sin^2\theta(1-e^2)+a^2e^2+a^2\cos^2\theta-2a^2e\cos\theta$$ $$=a^2(\sin^2\theta-e^2\sin^2\theta+e^2+\cos^2\theta-2e\cos\theta)=a^2(1+e^2\cos^2\theta-2e\cos\theta)$$Hence we get $PF=a(1-e\cos\theta)$

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  • $\begingroup$ Hmmm ... Draw a circle of radius $|OF|$ about $O$ and say that it meets $OQ$ at $X$; and drop a perpendicular from $X$ to $X^\prime$ on the ellipse's major axis. Then $|OX^\prime| = c \cos\theta$, so that (writing $A$ for the ellipse's far-right vertex) $|AX^\prime| = a - c\cos\theta = a ( 1 - e\cos\theta) = |PF|$. Computations aside, I wonder: Is there a geometrically-obvious reason why $\overline{AX^\prime}\cong\overline{PF}$? I'm not seeing one (yet). $\endgroup$ – Blue Mar 23 '15 at 9:50
  • $\begingroup$ @Blue your point being...? $\endgroup$ – najayaz Mar 23 '15 at 13:35
  • $\begingroup$ Just "thinking out loud". It's interesting to me that the segments $\overline{AX^\prime}$ and $\overline{PF}$ are congruent, so I'm curious about whether there's a nice way to demonstrate this fact without resorting to calculating their actual lengths. The congruence simply strikes me as being more than an algebraic coincidence. Maybe there's a cool application of the reflection property at work here ... or something. I may or may not have time to investigate this myself, so I thought I'd post the idea in case it might pique someone else's curiosity. $\endgroup$ – Blue Mar 23 '15 at 13:58

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