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MHT CET · Maths · Ellipse

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

  1. A \(\frac{\sqrt{3}}{2}\)
  2. B \(\frac{1}{\sqrt{3}}\)
  3. C \(\frac{1}{2}\)
  4. D \(\frac{1}{\sqrt{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{\sqrt{3}}{2}\)

Step-by-step Solution

Detailed explanation

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