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The ellipses have their centers at the origin and their major axes on the $x$-axis. Find the equation:

  1. with distance between directrices $27$, and between foci $3$;
  2. with a focus at $(-\sqrt{13},0)$ and a vertex at $(0,2)$.
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Welcome to MSE! Have you tried anything and can share your approach? Regards – Amzoti May 16 '13 at 3:19
@lexx, are they two independent questions? – lab bhattacharjee May 16 '13 at 3:27
@labbhattacharjee yes – lexx May 16 '13 at 5:34
@lexx, please find the answer – lab bhattacharjee May 16 '13 at 8:27

The equation of an ellipse with the center at the origin and the major axes on the x-axis is $$\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$$

where $2a,2b$ are the major & minor axes respectively.

We know the coordinate of the foci are $(\pm ae,0)$ and the equation of directrices are $x=\pm\frac ae$ where $e$ is the Eccentricity $e=\frac{\sqrt{a^2-b^2}}b$

So the distance between the foci is $2ae\implies 2ae=3$ and

the distance between the directrices is $\frac{2a}e\implies \frac{2a}e=27$

Solve for $a,e$

$2.$ As the coordinate of the foci are $(\pm ae,0), ae=\sqrt{13}$ as $ae>0$

The vertices are $(\pm a,0)$ and $(0,\pm b)\implies b=2$ as $b>0$

As $b^2=a^2(1-e^2)\iff a^2=b^2+(ae)^2=2^2+13=17$

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