Q86 / 150 · MHT CET 2020 (16 Oct Shift 2) · Mathematics

The eccentricity of the ellipse $y^{2}+4 x^{2}-12 x+6 y+14=0$ is

  • Correct answer
    A
    $\frac{\sqrt{3}}{2}$
  • B
    $\frac{1}{\sqrt{3}}$
  • C
    $\frac{1}{2}$
  • D
    $\frac{1}{\sqrt{2}}$

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, $a^{2}=1, b^{2}=4$ $\begin{array}{l} \therefore e=\sqrt{1-\frac{a^{2}}{b^{2}}} \\ e=\sqrt{1-\frac{1}{4}} \end{array}$

Hence, the correct answer is option: 'A'.

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