AP EAMCET · Maths · Hyperbola
Let \(\mathrm{e}\) be the eccentricity of the ellipse \(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{9}=1\). If \(\frac{1}{\mathrm{e}}\) is the eccentricity of a hyperbola, then the eccentricity of its conjugate hyperbola is
- A \(\frac{4}{3}\)
- B \(\frac{3}{\sqrt{5}}\)
- C \(\frac{4}{\sqrt{5}}\)
- D \(\frac{3}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
Given ellipse \(\frac{x^2}{4}+\frac{y^2}{9}=1\) Since \(b>a\) i.e. \(9>4\). Hence \[ \begin{aligned} & \Rightarrow e=\frac{1}{b} \sqrt{h^2-a^2}=\frac{1}{3} \sqrt{9-4} \\ & \Rightarrow e=\frac{\sqrt{5}}{3} \end{aligned} \] Let \(e_H=\frac{1}{e}=\frac{3}{\sqrt{5}}\) be the…
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