The equation of the ellipse whose foci are ( ±5 , 0 ) and one of its dierctrix is 5x = 36 , is

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Solution

The correct option is A

Foci ( ± 5 , 0 ) $\equiv$ ( ± ae , 0) . Directrix $(x = \frac{36}{5}) \equiv x = \frac{a}{e}$

So , $\frac{a}{e} = \frac{36}{5}, a e = 5 \Rightarrow a = 6 a n d e = \frac{5}{6}$

Therefore , b = 6 $\sqrt{1 - \frac{25}{36}} = 6 \frac{\sqrt{11}}{6} = \sqrt{11}$

Hence, the equation is $\frac{x^{2}}{36} + \frac{y^{2}}{11} = 1$

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