MHT CET · Maths · Ellipse
The eccentricity of the ellipse given by the equation \(9 x^{2}+16 y^{2}=144\) is
- A \(\frac{\sqrt{7}}{4}\)
- B \(\frac{1}{4}\)
- C \(\frac{\sqrt{3}}{4}\)
- D \(\frac{\sqrt{5}}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{7}}{4}\)
Step-by-step Solution
Detailed explanation
Given equation of an ellipse is
\(9 \mathrm{x}^{2}+16 \mathrm{y}^{2}=144 \)
\( \Rightarrow \frac{9 \mathrm{x}^{2}}{144}+\frac{16 \mathrm{y}^{2}}{144}=\frac{144}{144} \)
\( \Rightarrow \frac{\mathrm{x}^{2}}{16}+\frac{\mathrm{y}^{2}}{9}=1\)
Here, \(a^{2}=16, b^{2}=9 \quad\left[\because a^{2}>b^{2}\right] \)
\(\therefore\) Eccentricity \((e)=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}\)
\(9 \mathrm{x}^{2}+16 \mathrm{y}^{2}=144 \)
\( \Rightarrow \frac{9 \mathrm{x}^{2}}{144}+\frac{16 \mathrm{y}^{2}}{144}=\frac{144}{144} \)
\( \Rightarrow \frac{\mathrm{x}^{2}}{16}+\frac{\mathrm{y}^{2}}{9}=1\)
Here, \(a^{2}=16, b^{2}=9 \quad\left[\because a^{2}>b^{2}\right] \)
\(\therefore\) Eccentricity \((e)=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}\)
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