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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

Let \(P\) be a point on the ellipse \(\frac{x^2}{9}+\frac{y^2}{4}=1\). Let the line passing through \(P\) and parallel to \(y\)-axis meet the circle \(x^2+y^2=9\) at point \(Q\) such that \(P\) and \(Q\) are on the same side of the \(x\)-axis. Then, the eccentricity of the locus of the point \(R\) on \(P Q\) such that \(P R: R Q=4: 3\) as \(P\) moves on the ellipse, is :

  1. A  \(\frac{11}{19}\)
  2. B  \(\frac{13}{21}\)
  3. C \(\frac{\sqrt{139}}{23}\)
  4. D \(\frac{\sqrt{13}}{7}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\sqrt{13}}{7}\)

Step-by-step Solution

Detailed explanation

\( \mathrm{h}=3 \cos \theta \) \( \mathrm{k}=\frac{18}{7} \sin \theta \) \( \therefore \text { locus }=\frac{\mathrm{x}^2}{9}+\frac{49 \mathrm{y}^2}{324}=1 \) \( \mathrm{e}=\sqrt{1-\frac{324}{49 \times 9}}=\frac{\sqrt{117}}{21}=\frac{\sqrt{13}}{7}\)