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:
- $\frac{11}{19}$
- $\frac{13}{21}$
- $\frac{\sqrt{139}}{23}$
- $\frac{\sqrt{13}}{7}$
The Correct Option is D
Approach Solution - 1
\textbf{Given:} We are given the points: \begin{itemize} \item \( P (3C, 2S) \) \item \( Q (3C, 3S) \) \item \( R (h, k) \) \end{itemize} \textbf{Step 1: Express \( h \) and \( k \) in terms of \( \theta \)} The values for \( h \) and \( k \) are given as: \[ h = 3 \cos \theta \] \[ k = \frac{18}{7} \sin \theta \] \textbf{Step 2: Find the equation of the locus} We are given that the relationship between \( x \) and \( y \) for the locus is: \[ \frac{x^2}{9} + \frac{49y^2}{324} = 1 \] This equation represents an ellipse. \textbf{Step 3: Find the eccentricity of the ellipse} The eccentricity \( e \) of the ellipse is calculated as: \[ e = \sqrt{1 - \frac{324}{49 \times 9}} = \sqrt{\frac{117}{21}} = \frac{\sqrt{13}}{7} \]
Approach Solution -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: \(e = \frac{\sqrt{13}}{7}.\)
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