The eccentricity of the ellipse $4 x^2+25 y^2=100$ is
The eccentricity of the ellipse $4 x^2+25 y^2=100$ is
- $\frac{\sqrt{21}}{5}$
- $\frac{\sqrt{21}}{2}$
- $\frac{\sqrt{21}}{4}$
- $\frac{\sqrt{21}}{25}$
Solution
$4 x^2+25 y^2=100$
$
\frac{x^2}{25}+\frac{y^2}{4}=1
$
$\therefore a^2=25$ and $b^2=4$
Ecentricity $=\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{25-4}{25}}=\frac{\sqrt{21}}{5}$
Hence, option (1) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 2)
Practice more Conic Sections questions on Aicharya