KCET · Maths · Ellipse
The eccentricity of the ellipse \( 9 x^{2}+25 y^{2}=225 \) is
- A \( \frac{3}{5} \)
- B \( \frac{9}{16} \)
- C \( \frac{4}{5} \)
- D \( \frac{3}{4} \)
Answer & Solution
Correct Answer
(C) \( \frac{4}{5} \)
Step-by-step Solution
Detailed explanation
(C)
\(9 x^{2}+25 y^{2}=225\)
i.e., \(\frac{9 x^{2}}{225}+\frac{25 y^{2}}{225}=1\)
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
\(\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1\)
\(a=5, b=3\)
\(c=\sqrt{a^{2}-b^{2}}=\sqrt{5^{2}-3^{2}}=\sqrt{16}=4\)
Eccentricity \(e=\frac{C}{a}=\frac{4}{5}\)
\(9 x^{2}+25 y^{2}=225\)
i.e., \(\frac{9 x^{2}}{225}+\frac{25 y^{2}}{225}=1\)
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
\(\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1\)
\(a=5, b=3\)
\(c=\sqrt{a^{2}-b^{2}}=\sqrt{5^{2}-3^{2}}=\sqrt{16}=4\)
Eccentricity \(e=\frac{C}{a}=\frac{4}{5}\)
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