Q97 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics

If the product of the lengths of the perpendiculars drawn from the foci to the tangent $y=\frac{-3}{4} x+3 \sqrt{2}$ of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is 9 , then the eccentricity of that ellipse is

  • A
    $\frac{\sqrt{2}}{3}$
  • B
    $\frac{\sqrt{5}}{6}$
  • C
    $\frac{1}{9}$
  • Correct answer
    D
    $\frac{\sqrt{7}}{4}$

Solution

We know that, the product of the length of the perpendicular drawn from the foci to the tangent of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $b^2$. Also, length $=9 \quad \therefore \quad b^2=9$ Equation of tangent $y=\frac{-3}{4} x+3 \sqrt{2}$ $\therefore \quad 3 \sqrt{2}= \pm \sqrt{a^2\left(\frac{-3}{4}\right)^2+b^2}$ $18=\frac{9}{16} a^2+9 \Rightarrow 2-1=\frac{a^2}{16} \Rightarrow a^2=16$ $\therefore \quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$

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

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