What is the value of $SA\cdot S'A$ for the hyperbola ${ x }^{ 2 }-2{ y }^{ 2 }=1$?

Question

$S,S'$ are the foci and $A$ is the vertex. I have found the eccentricity $e$ to be $\sqrt { \frac { 3 }{ 2 } } $ . I have also found out $S\equiv (\sqrt { \frac { 3 }{ 2 } } ,0)$, $S'\equiv (-\sqrt { \frac { 3 }{ 2 } } ,0)$ and $A\equiv (1,0)$. How do I proceed?

Answer 1

\begin{align*} \frac{x^2}{a^2}-\frac{y^2}{b^2} &= 1 \\ e &= \frac{\sqrt{a^2+b^2}}{a} \\ SA \times S'A &= (ae-a)(ae+a) \\ &= a^2(e^2-1) \\ &= b^2 \\ &= \frac{1}{2} \end{align*}

For more information on product properties of confocal conics, see another answer here.