Question:
The eccentricity of the hyperbola \(\frac{(x-3)^2}{9}+\frac{4(y-1)^2}{45}=1\) is equal to
The eccentricity of the hyperbola \(\frac{(x-3)^2}{9}+\frac{4(y-1)^2}{45}=1\) is equal to
Updated On: Jun 8, 2024
- \(\frac{3}{\sqrt{5}}\)
- \(\frac{5}{3}\)
- \(\frac{5}{\sqrt{3}}\)
- \(\frac{5}{2}\)
- \(\frac{3}{2}\)
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The Correct Option is
Solution and Explanation
The correct option is (E) : \(\frac{3}{2}\)
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