Q100 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Mathematics

Let $e_1$ be the eccentricity of a hyperbola for which distance between its focii is 2 times the distance between its directrices and $e_2$ be the eccentricity of another hyperbola for which the length of its transverse axis is twice the length of its the conjugate axis. Then $e_1 e_2=$

  • A
    1
  • Correct answer
    B
    $\frac{\sqrt{10}}{2}$
  • C
    $\sqrt{5}$
  • D
    $\frac{\sqrt{5}}{2}$

Solution

Given $e_1$ is the eccentricity of hyperbola and eccentricity $\mathrm{e}_2$ of another hyperbola. According to question, $$ \begin{aligned} & 2 \mathrm{a}_1 \mathrm{e}_1=2 \times \frac{2 \mathrm{a}_1}{\mathrm{e}_1} \\ & \mathrm{e}_1^2=2 \\ & \mathrm{e}_1= \pm \sqrt{2} \end{aligned} $$ Here, $e_1>1$ then $e_1=\sqrt{2}$ Now, $2 \mathrm{a}_2=4 \mathrm{~b}_2$ $$ \begin{aligned} & \mathrm{a}_2=2 \mathrm{~b}_2 \\ & \mathrm{~b}_2=\frac{\mathrm{a}_2}{2} \end{aligned} $$ So, $c_2^2=a_2^2+b_2^2$ $$ \begin{aligned} & \mathrm{c}_2^2=\mathrm{a}_2^2+\frac{\mathrm{a}_2^2}{4}=\frac{5 \mathrm{a}_2^2}{4} \\ & \mathrm{c}_2=\frac{\sqrt{5}}{2} \mathrm{a}_2 \\ & \frac{\mathrm{c}_2}{\mathrm{a}_2}=\frac{\sqrt{5}}{2} \\ & \mathrm{e}_2=\frac{\sqrt{5}}{2} \end{aligned} $$ Then, $\mathrm{e}_1 \mathrm{e}_2=\sqrt{2} \times \frac{\sqrt{5}}{2}=\frac{\sqrt{10}}{2}$ So, option (b) is correct.

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

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