Let P be a point on the ellipse `(x^(2))/(9)+(y^(2))/(4)=1` and the line through P parallel to the y-axis meets the circle `x^(2)+y^(2)=9` at Q where P,Q are on the same side of the x-axis. If R is a point on PQ such that `(PR)/(RQ)=(1)/(2)`, then locus of R is

Options

A. `(x^(2))/(9)+(9y^(2))/(49)=1`
B. `(x^(2))/(49)+(y^(2))/(9)=1`
C. `(x^(2))/(9)+(y^(2))/(49)=1`
D. `(9x^(2))/(49)+(y^(2))/(49)=1`

Answer

To solve the problem step by step, we will follow the given information and derive the locus of point R.

### Step 1: Identify the coordinates of point P on the ellipse
The equation of the ellipse is given by:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
The coordinates of any point P on this ellipse can be expressed as:
\[ P(3 \cos \theta, 2 \sin \theta) \]
where \( \theta \) is a parameter.

### Step 2: Find the coordinates of point Q on the circle
The equation of the circle is given by:
\[ x^2 + y^2 = 9 \]
The coordinates of any point Q on this circle can be expressed as:
\[ Q(3 \cos \phi, 3 \sin \phi) \]
where \( \phi \) is another parameter.

### Step 3: Determine the relationship between P and Q
Since the line through P is parallel to the y-axis, the x-coordinates of P and Q must be equal. Therefore, we have:
\[ 3 \cos \theta = 3 \cos \phi \implies \cos \theta = \cos \phi \]
This implies that \( \phi = \theta \) or \( \phi = -\theta \). Since P and Q are on the same side of the x-axis, we will take \( \phi = \theta \).

### Step 4: Find the coordinates of point R using the section formula
Point R divides the line segment PQ in the ratio \( PR : RQ = 1 : 2 \). According to the section formula, the coordinates of point R can be calculated as follows:
\[ R\left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}\right) \]
where \( m = 1 \), \( n = 2 \), \( P(3 \cos \theta, 2 \sin \theta) \), and \( Q(3 \cos \theta, 3 \sin \theta) \).

Thus, the coordinates of R are:
\[ R\left(\frac{1 \cdot 3 \cos \theta + 2 \cdot 3 \cos \theta}{1 + 2}, \frac{1 \cdot 3 \sin \theta + 2 \cdot 2 \sin \theta}{1 + 2}\right) \]
\[ R\left(\frac{3 \cos \theta + 6 \cos \theta}{3}, \frac{3 \sin \theta + 4 \sin \theta}{3}\right) \]
\[ R\left(3 \cos \theta, \frac{7 \sin \theta}{3}\right) \]

### Step 5: Replace \( \cos \theta \) and \( \sin \theta \) with \( x \) and \( y \)
Let \( x = 3 \cos \theta \) and \( y = \frac{7}{3} \sin \theta \). From the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we have:
\[ \left(\frac{y}{\frac{7}{3}}\right)^2 + \left(\frac{x}{3}\right)^2 = 1 \]
This simplifies to:
\[ \frac{9y^2}{49} + \frac{x^2}{9} = 1 \]

### Final Step: Write the locus equation
The locus of point R is given by:
\[ \frac{x^2}{9} + \frac{9y^2}{49} = 1 \]

### Conclusion
Thus, the locus of point R is:
\[ \frac{x^2}{9} + \frac{9y^2}{49} = 1 \]