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AP EAMCET · Maths · Ellipse

The eccentricity of the ellipse \(4 x^2+25 y^2=100\) is

  1. A \(\frac{\sqrt{21}}{5}\)
  2. B \(\frac{\sqrt{21}}{2}\)
  3. C \(\frac{\sqrt{21}}{4}\)
  4. D \(\frac{\sqrt{21}}{25}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{\sqrt{21}}{5}\)

Step-by-step Solution

Detailed explanation

\(4 x^2+25 y^2=100\) \[ \frac{x^2}{25}+\frac{y^2}{4}=1 \] \(\therefore a^2=25\) 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.