Let P be a point on the ellipse x^2 / 9 + y^2 / 4 = 1. ​​​​ Let the line passing through P and parallel to y-axis meet the circle ​

Question

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

(1) \(\frac{11}{19}\)

(2) \(\frac{13}{21}\)

(3) \(\frac{\sqrt{139}}{23}\)

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

Answer 1

Correct option is (4) \(\frac{\sqrt{13}}{7}\)

\(\mathrm{h}=3 \cos \theta\)

\(\mathrm{k}=\frac{18}{7} \sin \theta\)

\(\therefore \text{locus }=\frac{\mathrm{x}^{2}}{9}+\frac{49 \mathrm{y}^{2}}{324}=1\)

\(\mathrm{e}=\sqrt{1-\frac{324}{49 \times 9}}=\frac{\sqrt{117}}{21}=\frac{\sqrt{13}}{7}\)