# Ellipse Proof

• Sep 8th 2012, 06:06 AM
lanierms
Ellipse Proof
Attachment 24734 Use the following ellipse : $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (F, F' are foci of the ellipse and the line passes through it.)

Prove : $\displaystyle \frac{1}{PF}+\frac{1}{QF} = \frac{2a}{b^2}$

I tried to prove it using directrix, but it got way too complicated.
• Sep 8th 2012, 11:15 AM
MaxJasper
Re: Ellipse Proof
Quote:

Originally Posted by lanierms
Attachment 24734 Use the following ellipse : $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (F, F' are foci of the ellipse and the line passes through it.)

Prove : $\displaystyle \frac{1}{PF}+\frac{1}{QF} = \frac{2a}{b^2}$

I tried to prove it using directrix, but it got way too complicated.

If the equality is true for all points P & Q on ellipse then letting P(a,0) & Q(-a,0) we will have:

$\displaystyle e\text{:=}\sqrt{1-\frac{b^2}{a^2}}$

$\displaystyle \frac{1}{\text{PF}}+\frac{1}{\text{QF}} =\frac{1}{a-ea}+\frac{1}{a+ea}= \frac{2 a}{b^2}$

In the general case when P & Q are elsewhere on the ellipse. We have:

Move x-y coordinate's origin to focus F...then PF & QF will be:

$\displaystyle \text{PF}\text{:=}a(1-e \text{Cos}[\theta ])$
$\displaystyle \text{QF}\text{:=}a(1-e \text{Cos}[\theta -\pi ])$

$\displaystyle \frac{1}{\text{PF}}+\frac{1}{\text{QF}} = \frac{2}{a-a e^2 \text{Cos}[\theta ]^2} = -\frac{2}{-a+\left(a-\frac{b^2}{a}\right) \text{Cos}[\theta ]^2}$
• Sep 8th 2012, 04:31 PM
lanierms
Re: Ellipse Proof
It seems like a nice proof, but can there be another way without using eccentricity?
• Sep 9th 2012, 04:27 PM
MaxJasper
Re: Ellipse Proof
Variation of $\displaystyle \frac{1}{\text{PF}}+\frac{1}{\text{QF}}$ by $\displaystyle \theta$ fpr a=3, b=2

http://mathhelpforum.com/attachment....1&d=1347236778
• Sep 10th 2012, 07:49 AM
lanierms
Re: Ellipse Proof
I don't quite understand the part where you said
PF = a(1-ecos(theta))