Question

The eccentricity of the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$ isGiven, equation of the ellipse frac $x^2 36+\operatorname{frac} y^2 16=1$

• A. $\frac{\sqrt{5}}{3}$
• B. $\frac{\sqrt{5}}{6}$
• C. $\frac{\sqrt{30}}{6}$
• D. $\frac{\sqrt{10}}{6}$
• E. $\frac{\sqrt{30}}{7}$

Answer 1

Answer: (A)

Given, equation of the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$
Here, $a^2=36, b^2=16$
We know that,
$ b^2=a^2\left(1-e^2\right) $
$ \Rightarrow 16=36\left(1-e^2\right) $
$\Rightarrow 1-e^2=\frac{16}{36}=\frac{4}{9}$
$\Rightarrow e^2=1-\frac{4}{9} \Rightarrow e^2=\frac{5}{9}$
$ \Rightarrow e=\frac{\sqrt{5}}{3}$