1. ## Ellipse proofs

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 $PG^2+(1-e^2)PS.PS'$
where e is the eccentricity of the ellipse.
I let P(s,t)
the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
where $b^2=a^2(1-e^2) \rightarrow(1-e^2)=\frac{b^2}{a^2}$
S(ae,0) and s'(-ae,0)
gradient of normal at P= $-\frac{2a^2y}{b^2(a^2-2x)}$
equation PG $y-t=-\frac{2a^2t}{b^2(a^2-2s)}(x-s)$
when y=0 $x=s+\frac{b^2}{a^2}\frac{a^2-2s}{2}$
then
$PG^2=(s-(s+\frac{b^2}{a^2}\frac{a^2-2s}{2}))^2+t^2$
$=(\frac{b^2}{a^2})^2(\frac{(a^2-2s)^2}{4})+t^2$
$=(1-e^2)^2(\frac{(a^2-2s)^2}{4})+t^2$

PS.PS'= $\sqrt{(s-ae)^2+t^2}\times\sqrt{(s+ae)^2+t^2}$
Now my problem is here, i cannot find a square root for this expression.
Thanks

2. First of all let us see what will be PS*PS'.
PS = e*PM, where PM is the distance of P from the diretrix.
PS' = e*PM'. If co-ordinates of P are (x1, y1),
PM = (a/e - x1) and PM' = (a/e + x1). Hence
PS*PS' = e^2(a^2/e^2 - x1^2) = (a^2 - e^2*x1^2)........(1)
The equation of the tangent to the ellipse is given by
xx1/a^2 + yy1/b^2 = 1
Slope of the tangent is -x1*b^2/y1*a^2
Slope of the normal is y1a^2/x1b^2
Find the equation of the normal.
Find the point of intersection of the normal with x-axis.
Find PG^2.
Replace y1^2 in the above expression by b^2(1 - x^2/a^2)
Now simplify.