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An ellipse has the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a>b$, and with eccentricity $e$. It also has foci $S$ and $S'$ and directrices $l$ and $l'$.

a) Use the focus-directrix property to show that $PS+PS'=2a$, where $P$ is a point on the ellipse.

b) Hence find an expression involving $a$ and $e$ for the perimeter of triangle $SPS'$.

c) In the case that angle $SPS'=90$ degrees, show that $(PS)(PS')=2b^2$.

I did parts a and b easily enough, but I can't do part c. For part b I got $2a(1+e)$.

Can someone help me with part c?

Mathmo123
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Ahmed
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1 Answers1

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Let $x$ be the length of $PS'$, so that the length of $PS$ will be $2a-x$, as you have shown that $PS'+PS=2a$. The objective is to find $(PS)(PS')=\color{red}{(2a-x)x}$ for when $\angle SPS'=90^\circ$.

You have also worked out the perimeter of $SPS'$, which is $2a(1+e)$, where the eccentricity $e=\sqrt{1-\frac{b^2}{a^2}}$.

Thus, the hypotenuse of $SPS'$, which is side $SS'$ will be $2a(1+e)-2a=2ae$.

As we have a right angled triangle, $SPS'$, we can use Pythagoras theorem:- $$ |PS'|^2+|PS|^2=|SS'|^2\\\Rightarrow x^2+(2a-x)^2=(2ae)^2\\\Rightarrow x^2+4a^2-4ax+x^2=4a^2\left(1-\frac{b^2}{a^2}\right)$$ Can you take it from here?

Alijah Ahmed
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