Q97 / 160 · AP EAMCET 2023 (19 May Shift 1) · Mathematics

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
    $\frac{4}{\sqrt{5}}$
  • Correct answer
    D
    $\frac{3}{2}$

Solution

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 eccentricity of hyperbola and $e_C$ be the eccentricity of the conjugate hypérbola. We know that, for a hyperbola- $$ \begin{aligned} & \Rightarrow \quad \frac{1}{e_H^2}+\frac{1}{e_C^2}=1 \Rightarrow\left(\frac{\sqrt{5}}{3}\right)^2+\frac{1}{e_C^2}=1 \\ & \Rightarrow e_C=3 / 2 \end{aligned} $$

Hence, the correct answer is option: 'D'.

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