Let P be a point on the ellipse , Let the line passing through P and parallel to y-axis meet the circle 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 and PQ such that PR : RQ = 4 : 3 a

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Solution

Q.
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 and PQ such that PR : RQ = 4 : 3 as P moves on the ellipse, is [2024]

1 $\frac{13}{21}$

2 $\frac{\sqrt{139}}{23}$

3 $\frac{\sqrt{13}}{7}$

4 $\frac{11}{19}$

Ans.
(3)

As, P be a point on the ellipse.

$∴$ Coordinates of P = (3 cos $θ$ , 2 sin $θ$ )

Also, Q is a point on the circle $x^{2} + y^{2} = 9$

$∴$ Coordinates of Q = (3 cos $θ$ , 3 sin $θ$ )

Now, R(h, k) divides the line PQ in the ratio 4 : 3.

$∴ h = \frac{12 \cos θ + 9 \cos θ}{7} = \frac{21 \cos θ}{7} = 3 \cos θ$

and $k = \frac{12 \sin θ + 6 \sin θ}{7} = \frac{18}{7} \sin θ$

Now, $\frac{h^{2}}{9} + \frac{49 k^{2}}{324} = 1$

So, locus of R, $\frac{x^{2}}{3^{2}} + \frac{y^{2}}{{(18 / 7)}^{2}} = 1$ is ellipse.

$∴$ Eccentricity = $\sqrt{1 - \frac{{(18 / 7)}^{2}}{9}} = \sqrt{\frac{13}{49}} = \frac{\sqrt{13}}{7}$ .

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