P is a point on ellipse `x^2/9+y^2/4=1`and a line through 'P' parallel to y-axes intersect its auxiliary circle on 94 the same side of major axis at Q, then eccentricity of locus of point 'R' which divides PQ internally in the ratio 4: 3 is

Options

A. `sqrt13/7`
B. `sqrt20/7`
C. `(2sqrt5)/7`
D. `sqrt5/3`

Answer

To solve the problem step by step, we will follow the given information about the ellipse and the auxiliary circle, and then find the eccentricity of the locus of point R.

### Step-by-Step Solution:

1. **Identify the Ellipse and its Auxiliary Circle:**
The given ellipse is represented by the equation:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
Here, \(a^2 = 9\) and \(b^2 = 4\), which gives us \(a = 3\) and \(b = 2\). The auxiliary circle of the ellipse has a radius equal to \(a\) and is centered at the origin, given by:
\[ x^2 + y^2 = 9 \]

2. **Point P on the Ellipse:**
Let point \(P\) on the ellipse be represented in parametric form:
\[ P = (3\cos\theta, 2\sin\theta) \]

3. **Finding Point Q:**
A line through point \(P\) that is parallel to the y-axis will have the same x-coordinate as point \(P\). Therefore, the equation of this line is:
\[ x = 3\cos\theta \]
To find point \(Q\), we substitute \(x = 3\cos\theta\) into the equation of the auxiliary circle:
\[ (3\cos\theta)^2 + y^2 = 9 \]
Simplifying this gives:
\[ 9\cos^2\theta + y^2 = 9 \implies y^2 = 9 - 9\cos^2\theta \implies y^2 = 9(1 - \cos^2\theta) = 9\sin^2\theta \]
Thus, \(y = 3\sin\theta\). Therefore, point \(Q\) is:
\[ Q = (3\cos\theta, 3\sin\theta) \]

4. **Finding Point R:**
Point \(R\) divides segment \(PQ\) in the ratio \(4:3\). Using the section formula, the coordinates of point \(R\) are given by:
\[ R = \left(\frac{4 \cdot 3\cos\theta + 3 \cdot 3\cos\theta}{4 + 3}, \frac{4 \cdot 2\sin\theta + 3 \cdot 3\sin\theta}{4 + 3}\right) \]
Simplifying this gives:
\[ R = \left(\frac{12\cos\theta + 9\cos\theta}{7}, \frac{8\sin\theta + 9\sin\theta}{7}\right) = \left(\frac{21\cos\theta}{7}, \frac{17\sin\theta}{7}\right) \]
Thus, we have:
\[ R = (3\cos\theta, \frac{17}{7}\sin\theta) \]

5. **Finding the Locus of R:**
To find the locus of point \(R\), we eliminate \(\theta\). We know:
\[ h = 3\cos\theta, \quad k = \frac{17}{7}\sin\theta \]
From \(h\), we have \(\cos\theta = \frac{h}{3}\). Substituting into the equation for \(k\):
\[ k = \frac{17}{7}\sqrt{1 - \left(\frac{h}{3}\right)^2} \]
Squaring both sides gives:
\[ k^2 = \left(\frac{17}{7}\right)^2 \left(1 - \frac{h^2}{9}\right) \]
Rearranging leads to:
\[ \frac{h^2}{9} + \frac{7k^2}{17^2} = 1 \]

6. **Identifying the Eccentricity:**
The locus is an ellipse with semi-major axis \(a = 3\) and semi-minor axis \(b = \frac{17}{7}\). The eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \left(\frac{b}{a}\right)^2} \]
Substituting the values:
\[ e = \sqrt{1 - \left(\frac{17/7}{3}\right)^2} = \sqrt{1 - \left(\frac{17}{21}\right)^2} = \sqrt{1 - \frac{289}{441}} = \sqrt{\frac{152}{441}} = \frac{\sqrt{152}}{21} \]

### Final Answer:
The eccentricity of the locus of point \(R\) is:
\[ e = \frac{\sqrt{152}}{21} \]