Ellipse Proof

**lanierms** · September 8th 2012, 06:06 AM

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.

**MaxJasper** · September 8th 2012, 11:15 AM

Originally Posted by lanierms

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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:

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

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