Question

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

• 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,
$b^2=a^2\left(1−e^2\right)$
$⇒16=36\left(1−e^2\right)$
$⇒1−e^2=\frac{16}{36}=\frac{4}{9}$
$⇒e^2=1−\frac{4}{9}⇒e^2=\frac{5}{9}$
$⇒e=\frac{\sqrt{5}}{3}$