$S,S'$ 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?
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\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.
Ng Chung Tak
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$SA . S'A = (ae-a)(ae+a)$ How did you get this ? How is $SA = ae -a$ ? – Gaurav Lakhotia Mar 08 '17 at 09:08
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For standard form, vertices are $A(a,0)$, $A'(-a,0)$ and foci are $S(ae,0)$, $S'(-ae,0)$. Therefore, $SA=(ae-a,0)$ and $S'A=(ae+a,0)$. – Ng Chung Tak Mar 08 '17 at 09:11