Q85 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics

The eccentricity of the ellipse $4 x^2+25 y^2=100$ is

  • Correct answer
    A
    $\frac{\sqrt{21}}{5}$
  • B
    $\frac{\sqrt{21}}{2}$
  • C
    $\frac{\sqrt{21}}{4}$
  • D
    $\frac{\sqrt{21}}{25}$

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.

Hence, the correct answer is option: 'A'.

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