Question:
The eccentricity of the hyperbola $\frac{(x-1)^{2}}{25}-\frac{(y+2)^{2}}{11}=1$ is:
The eccentricity of the hyperbola $\frac{(x-1)^{2}}{25}-\frac{(y+2)^{2}}{11}=1$ is:
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For hyperbolas, $e > 1$. If your result is less than 1, you likely calculated eccentricity for an ellipse.
Updated On: Apr 28, 2026
- $\frac{5}{3}$
- $\frac{6}{5}$
- $\frac{7}{5}$
- $\frac{25}{11}$
- $\frac{5}{11}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Step 2: Analysis
Comparing with the given equation, $a^2 = 25$ and $b^2 = 11$.
Step 3: Calculation
$e = \sqrt{1 + \frac{11}{25}} = \sqrt{\frac{25 + 11}{25}} = \sqrt{\frac{36}{25}} = \frac{6}{5}$. Final Answer: (B)
For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Step 2: Analysis
Comparing with the given equation, $a^2 = 25$ and $b^2 = 11$.
Step 3: Calculation
$e = \sqrt{1 + \frac{11}{25}} = \sqrt{\frac{25 + 11}{25}} = \sqrt{\frac{36}{25}} = \frac{6}{5}$. Final Answer: (B)
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