Let P be a point on the ellipse x^29 + y^24 = 1. Let the line pas… — Solution | JEE Mathematics

MCQMediumJEE 2024Conic Sections (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 the $y$ -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 PQ such that $PR : RQ = 4 : 3$ as P moves on the ellipse, is:

A
$\frac{11}{19}$
B
$\frac{13}{21}$
C
$\frac{\sqrt{139}}{23}$
D
$\frac{\sqrt{13}}{7}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Point P lies on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ . The vertical line through P meets the circle $x^2 + y^2 = 9$ at Q, with P and Q on the same side of the $x$ -axis. Point R divides PQ internally in the ratio $4:3$ .

Find: The eccentricity of the locus of R.

Take

P = (3\cos\theta, 2\sin\theta)

since $P$ lies on the ellipse.

Because the line through $P$ is parallel to the $y$ -axis, its equation is

x = 3\cos\theta

Substituting into the circle,

(3\cos\theta)^2 + y^2 = 9

y^2 = 9(1-\cos^2\theta) = 9\sin^2\theta

Since $P$ and $Q$ are on the same side of the $x$ -axis,

Q = (3\cos\theta, 3\sin\theta)

Using the section formula for $PR:RQ = 4:3$ ,

x_R = \frac{4x_Q + 3x_P}{7} = \frac{4(3\cos\theta) + 3(3\cos\theta)}{7} = 3\cos\theta

and

y_R = \frac{4y_Q + 3y_P}{7} = \frac{4(3\sin\theta) + 3(2\sin\theta)}{7} = \frac{18}{7}\sin\theta

Hence,

R = \left(3\cos\theta, \frac{18}{7}\sin\theta\right)

Therefore the locus of $R$ is

\frac{x^2}{9} + \frac{y^2}{\left(\frac{18}{7}\right)^2} = 1

that is,

\frac{x^2}{9} + \frac{49y^2}{324} = 1

This is an ellipse with

a = 3, \qquad b = \frac{18}{7}

So its eccentricity is

e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{\left(\frac{18}{7}\right)^2}{9}} = \sqrt{1 - \frac{324}{441}} = \sqrt{\frac{117}{441}} = \frac{\sqrt{13}}{7}

Therefore, the eccentricity of the locus is $\frac{\sqrt{13}}{7}$ . The correct option is D.

Common mistakes

Using the wrong section formula. For $PR:RQ = 4:3$ , the coordinates of R are weighted as $\frac{4Q+3P}{7}$ , not by directly averaging or reversing the weights. A wrong ratio gives an incorrect locus.
Taking the wrong point on the circle. Since P and Q are on the same side of the $x$ -axis, the sign of the $y$ -coordinate of Q must match that of P. Choosing $Q=(3\cos\theta,-3\sin\theta)$ is incorrect.
Using the incorrect parametric point for the ellipse. For $\frac{x^2}{9}+\frac{y^2}{4}=1$ , the correct parametrization is $P=(3\cos\theta,2\sin\theta)$ . Replacing the coefficient of $\sin\theta$ incorrectly changes the entire locus.

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Answer

Step-by-step solution

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