The eccentricity of the ellipse $y^{2}+4 x^{2}-12 x+6 y+14=0$ is
The eccentricity of the ellipse $y^{2}+4 x^{2}-12 x+6 y+14=0$ is
- $\frac{\sqrt{3}}{2}$
- $\frac{1}{\sqrt{3}}$
- $\frac{1}{2}$
- $\frac{1}{\sqrt{2}}$
Solution
The eccentricity of the ellipse $y^{2}+4 x^{2}-12 x+6 y+14=0$ is $\frac{\sqrt{3}}{2}$.
Given ellipse,
$\begin{array}{l}
y^{2}+4 x^{2}-12 x+6 y+14=0 \\
4 x^{2}-12 x+y^{2}+6 y+14=0 \\
4\left(x^{2}-3 x+\frac{9}{4}\right)+y^{2}+6 y+9=-14+9+9 \\
4\left(x-\frac{3}{2}\right)^{2}+(y+3)^{2}=4 \\
\frac{\left(x-\frac{3}{2}\right)^{2}}{1}+\frac{(y+3)^{2}}{4}=1
\end{array}$
Here, $a^{2}=1, b^{2}=4$
$\begin{array}{l}
\therefore e=\sqrt{1-\frac{a^{2}}{b^{2}}} \\
e=\sqrt{1-\frac{1}{4}}
\end{array}$
Asked in: MHT CET 2020 (16 Oct Shift 2)
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