Question

The eccentricity of the ellipse $25 x^2+16 y^2=400$ is

(a) $\frac{3}{5}$

(b) $\frac{1}{3}$

(c) $\frac{2}{5}$

(d) $\frac{1}{5}$

Answer 1

SOLUTION — $\because \quad \frac{x^2}{16}+\frac{y^2}{25}=1$

Here ,$\begin{aligned}a^2 & =16, b^2=25 \\e & =\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}\end{aligned}$

So, The correct option of this question will be (A).