The foci of an ellipse are S, S' and P is any point on the curve. The normal at P meets SS' at G. Prove the $\displaystyle PG^2+(1-e^2)PS.PS'$

whereeis the eccentricity of the ellipse.

I let P(s,t)

the equation of the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

where $\displaystyle b^2=a^2(1-e^2) \rightarrow(1-e^2)=\frac{b^2}{a^2}$

S(ae,0) and s'(-ae,0)

gradient of normal at P=$\displaystyle -\frac{2a^2y}{b^2(a^2-2x)}$

equation PG $\displaystyle y-t=-\frac{2a^2t}{b^2(a^2-2s)}(x-s)$

when y=0 $\displaystyle x=s+\frac{b^2}{a^2}\frac{a^2-2s}{2}$

then

$\displaystyle PG^2=(s-(s+\frac{b^2}{a^2}\frac{a^2-2s}{2}))^2+t^2$

$\displaystyle =(\frac{b^2}{a^2})^2(\frac{(a^2-2s)^2}{4})+t^2$

$\displaystyle =(1-e^2)^2(\frac{(a^2-2s)^2}{4})+t^2$

PS.PS'=$\displaystyle \sqrt{(s-ae)^2+t^2}\times\sqrt{(s+ae)^2+t^2}$

Now my problem is here, i cannot find a square root for this expression.

Thanks