3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

enter image description here

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 $PF=a(1-e\cos\theta)$

$\endgroup$
  • $\begingroup$ 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...? $\endgroup$ – G-man 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.