Question

In an ellipse, if foci are $(\pm 5,0)$ and $x=\frac{36}{5}$ as one of the directrix, then the equation of the ellipse is

• A. $\frac{x^2}{36}+\frac{y^2}{11}=1$
• B. $\frac{x^2}{11}+\frac{y^2}{36}=1$
• C. $\frac{x^2}{121}+\frac{y^2}{36}=1$
• D. $\frac{x^2}{36}+\frac{y^2}{121}=1$

Answer 1

Answer: (A)

We have, $ae=5,\frac{a}{e}=\frac{36}{5}$
which give $a^2=36$ or $a=6$.
$\because$ directrix $x=\pm \frac{a}{e}$.
Therefore, $e=\frac{5}{6}$.
Now, $b=a\sqrt{1-e^2}=6\sqrt{1-\frac{25}{36}}=\sqrt{11}$.
Thus, the equation of the ellipse is $\frac{x^2}{36}+\frac{y^2}{11}=1$