Question

The eccentricity of the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$ is

• A. $\frac{2\sqrt{5}}{6}$
• B. $\frac{2\sqrt{5}}{4}$
• C. $\frac{2\sqrt{13}}{6}$
• D. $\frac{2\sqrt{13}}{4}$

Answer 1

Answer: (A)

$\frac{2\sqrt{5}}{6}$

Answer 2

We have
$\frac{x^2}{36} + \frac{y^2}{16} = 1$
$\Rightarrow \frac{x^2}{6^2} + \frac{y^2}{4^2}= 1$
$\therefore a=6$ and $b=4$
Since, $a > b$
$\therefore e=\sqrt{1-\frac{b^{2}}{a^{2}}}$
$=\sqrt{1-\frac{(4)^{2}}{(6)^{2}}}$
$=\sqrt{1-\frac{16}{36}}$
$=\sqrt{\frac{20}{36}}$
$=\frac{2 \sqrt{5}}{6}$