The equation of the ellipse whose foci are ( | vedclass

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Question

(a) Foci $( \pm 5,\,0) \equiv ( \pm ae,\,0)$.

Directrix $\left( {x = \frac{{36}}{5}} \right) \equiv x = \frac{a}{e}$

So, $\frac{a}{e} = \frac{{3

Vedclass · Accepted Answer

$\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{11}} = 1$

Vedclass · Answer

(a) Foci $( \pm 5,\,0) \equiv ( \pm ae,\,0)$.

Directrix $\left( {x = \frac{{36}}{5}} \right) \equiv x = \frac{a}{e}$

So, $\frac{a}{e} = \frac{{36}}{5},\;ae = 5$ ==> $a = 6$ and $e = \frac{5}{6}$

Therefore, $b = 6\sqrt {1 - \frac{{25}}{{36}}} = 6\frac{{\sqrt {11} }}{6} = \sqrt {11} $

Hence equation is $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{11}} = 1$.

The equation of the ellipse whose foci are $( \pm 5,\;0)$ and one of its directrix is $5x = 36$, is

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