Question:
The eccentricity of the ellipse
\[
y^2 + 4x^2 - 12x + 6y + 14 = 0
\]
is
The eccentricity of the ellipse
\[
y^2 + 4x^2 - 12x + 6y + 14 = 0
\]
is
Show Hint
To find the eccentricity of an ellipse, use the formula \( e = \sqrt{1 - \frac{b^2}{a^2}} \), where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.
Updated On: Mar 28, 2026
- \( \frac{\sqrt{3}}{2} \)
- \( \frac{1}{\sqrt{3}} \)
- \( \frac{1}{2} \)
- \( \frac{1}{\sqrt{2}} \)
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The Correct Option is A
Solution and Explanation
Step 1: Rewrite the equation of the ellipse.
We start by completing the square for both \( x \) and \( y \) terms in the equation: \[ y^2 + 4x^2 - 12x + 6y + 14 = 0 \] Rewrite the equation in standard form of the ellipse: \[ \frac{(x - 3)^2}{a^2} + \frac{(y + 1)^2}{b^2} = 1. \]
Step 2: Calculate the eccentricity.
The eccentricity \( e \) of an ellipse is given by the formula: \[ e = \sqrt{1 - \frac{b^2}{a^2}}. \] After solving, we find \( e = \frac{\sqrt{3}}{2} \).
Step 3: Conclusion.
Thus, the eccentricity of the ellipse is \( \frac{\sqrt{3}}{2} \), corresponding to option (A).
We start by completing the square for both \( x \) and \( y \) terms in the equation: \[ y^2 + 4x^2 - 12x + 6y + 14 = 0 \] Rewrite the equation in standard form of the ellipse: \[ \frac{(x - 3)^2}{a^2} + \frac{(y + 1)^2}{b^2} = 1. \]
Step 2: Calculate the eccentricity.
The eccentricity \( e \) of an ellipse is given by the formula: \[ e = \sqrt{1 - \frac{b^2}{a^2}}. \] After solving, we find \( e = \frac{\sqrt{3}}{2} \).
Step 3: Conclusion.
Thus, the eccentricity of the ellipse is \( \frac{\sqrt{3}}{2} \), corresponding to option (A).
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