The eccentricity of the ellipse given by the equation $9 x^{2}+16 y^{2}=144$ is
The eccentricity of the ellipse given by the equation $9 x^{2}+16 y^{2}=144$ is
- $\frac{\sqrt{7}}{4}$
- $\frac{1}{4}$
- $\frac{\sqrt{3}}{4}$
- $\frac{\sqrt{5}}{4}$
Solution
Given equation of an ellipse is
$\begin{array}{l}
9 \mathrm{x}^{2}+16 \mathrm{y}^{2}=144 \\
\Rightarrow \frac{9 \mathrm{x}^{2}}{144}+\frac{16 \mathrm{y}^{2}}{144}=\frac{144}{144} \\
\Rightarrow \frac{\mathrm{x}^{2}}{16}+\frac{\mathrm{y}^{2}}{9}=1
\end{array}$
$\begin{array}{l}
\text { Here, } a^{2}=16, b^{2}=9 \quad\left[\because a^{2}>b^{2}\right] \\
\therefore \text { Eccentricity }(e)=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4} .
\end{array}$
Asked in: MHT CET 2020 (12 Oct Shift 2)
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