1. ## Ellipse Proof

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

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

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

2. ## Re: Ellipse Proof

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

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

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:

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

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

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:

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

$\frac{1}{\text{PF}}+\frac{1}{\text{QF}} = \frac{2}{a-a e^2 \text{Cos}[\theta ]^2} = -\frac{2}{-a+\left(a-\frac{b^2}{a}\right) \text{Cos}[\theta ]^2}$

3. ## Re: Ellipse Proof

It seems like a nice proof, but can there be another way without using eccentricity?

4. ## Re: Ellipse Proof

Variation of $\frac{1}{\text{PF}}+\frac{1}{\text{QF}}$ by $\theta$ fpr a=3, b=2

5. ## Re: Ellipse Proof

I don't quite understand the part where you said
PF = a(1-ecos(theta))