Q96 / 160 · AP EAMCET 2023 (15 May Shift 2) · Mathematics

Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be equal to the length of its semi-major axis. If the radius of its director circle is $\sqrt{3}$ and $\mathrm{e}$ is its eccentricity, then the length of its latus rectum is

  • A
    $\frac{1}{\mathrm{a}}$
  • B
    $\frac{1}{b}$
  • Correct answer
    C
    $\frac{1}{\mathrm{e}}$
  • D
    $\frac{1}{a b}$

Solution

Given equation of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Length of latus rectum $=\frac{2 b^2}{a}=a$ $\Rightarrow 2 b^2=a^2...(i)$ and equation of director circle is $x^2+y^2=a^2+b^2 \Rightarrow x^2+y^2=3 b^2=(\sqrt{3} b)^2$ Since, $\sqrt{3} b=\sqrt{3} \Rightarrow b=1 \Rightarrow a=\sqrt{2}$ $\Rightarrow e=\frac{c}{a}=\frac{\sqrt{a^2-b^2}}{a}=\frac{b}{a}=\frac{1}{\sqrt{2}}$ Now 1. 1. $\mathrm{R}=a=\sqrt{2}=\frac{1}{e}$

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

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