The equation of the ellipse with focus at \(( \pm 5,0)\) and \(x=\frac{36}{5}\) as one directrix is
The equation of the ellipse with focus at \(( \pm 5,0)\) and \(x=\frac{36}{5}\) as one directrix is
\(\frac{x^2}{36}+\frac{y^2}{25}=1\)
\(\frac{x^2}{36}+\frac{y^2}{11}=1\)
\(\frac{x^2}{25}+\frac{y^2}{11}=1\)
None of these
Solution
We have ae \(=5\) [Since focus is \(( \pm \mathrm{ae}, 0)\)]
and \(\frac{\mathrm{a}}{\mathrm{e}}=\frac{36}{5} \quad\left[\right.\) since directrix is \(\left.\mathrm{x}= \pm \frac{\mathrm{a}}{\mathrm{e}}\right]\)
On solving we get \(\mathrm{a}=6\)
\(\text { and } \mathrm{e}=\frac{5}{6} \Rightarrow \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=36\left(1-\frac{25}{36}\right)=11\)
Thus, the required equation of the ellipse is
\(\frac{x^2}{36}+\frac{y^2}{11}=1\)