Question

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 1

The Correct Option is D

We are given the ellipse equation:
\(\frac{x^2}{9} + \frac{y^2}{24} = 1.\)
The general parametric equations for the ellipse are:
\(x = 3 \cos \theta, \quad y = 2 \sin \theta.\)
Thus, the coordinates of point \( P \) on the ellipse are \( P(3 \cos \theta, 2 \sin \theta) \).

\(\textbf{Step 1: Equation of the line passing through \( P \) and parallel to the y-axis.}\)

The line passing through \( P \) and parallel to the y-axis has the equation:
\(x = 3 \cos \theta,\)
since the x-coordinate is constant.

\(\textbf{Step 2: Finding the intersection point \( Q \) with the circle \( x^2 + y^2 = 9 \).}\)

Substitute \( x = 3 \cos \theta \) into the circle’s equation:

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

Simplifying:
\(y^2 = 9(1 - \cos^2 \theta) = 9 \sin^2 \theta.\)
Thus, the y-coordinate of point \( Q \) is \( y = 3 \sin \theta \), and the coordinates of \( Q \) are \( Q(3 \cos \theta, 3 \sin \theta) \).

\(\textbf{Step 3: Coordinates of the point \( R \) dividing \( PQ \) in the ratio \( PR : RQ = 4 : 3 \).}\)

We use the section formula to find the coordinates of \( R \). The coordinates of \( R \) dividing the line segment \( PQ \) in the ratio 4 : 3 are:
\(x_R = \frac{4x_Q + 3x_P}{4 + 3} = \frac{4(3 \cos \theta) + 3(3 \cos \theta)}{7} = \frac{21 \cos \theta}{7} = 3 \cos \theta,\)

\(y_R = \frac{4y_Q + 3y_P}{4 + 3} = \frac{4(3 \sin \theta) + 3(2 \sin \theta)}{7} = \frac{24 \sin \theta}{7} = \frac{24}{7} \sin \theta.\)
Thus, the coordinates of point \( R \) are \( R(3 \cos \theta, \frac{24}{7} \sin \theta) \).

\(\textbf{Step 4: Finding the eccentricity of the locus of point \( R \).}\)

The locus of \( R \) is given by:
\(\frac{x^2}{9} + \frac{y^2}{\left( \frac{24}{7} \right)^2} = 1.\)
Simplifying the second term:
\(\left( \frac{24}{7} \right)^2 = \frac{576}{49},\)
so the equation of the locus of \( R \) becomes:
\(\frac{x^2}{9} + \frac{49y^2}{576} = 1.\)
This is the equation of an ellipse with semi-major axis \( a = 3 \) and semi-minor axis \( b = \frac{24}{7} \).

The eccentricity \( e \) of an ellipse is given by:
\(e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{\left( \frac{24}{7} \right)^2}{9}} = \sqrt{1 - \frac{576}{441}} = \sqrt{\frac{441 - 576}{441}} = \sqrt{\frac{-135}{441}}.\)

Therefore, the eccentricity of the locus of point \( R \) is: \(e = \frac{\sqrt{13}}{7}.\)