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

Answer: (D)

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

Answer 2

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: $e = \frac{\sqrt{13}}{7}.$