AP EAMCET · Maths · Ellipse
The eccentricity of the ellipse \(4 x^2+25 y^2=100\) is
- A \(\frac{\sqrt{21}}{5}\)
- B \(\frac{\sqrt{21}}{2}\)
- C \(\frac{\sqrt{21}}{4}\)
- D \(\frac{\sqrt{21}}{25}\)
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.
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