P is any point on the auxililary circle of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` and Q is its corresponding point on the ellipse. Find the locus of the point which divides PQ in the ratio of `1:2`.

Answer

To find the locus of the point that divides the segment \( PQ \) in the ratio \( 1:2 \), where \( P \) is any point on the auxiliary circle of the ellipse given by

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

and \( Q \) is its corresponding point on the ellipse, we can follow these steps:

### Step 1: Identify Points on the Ellipse and Auxiliary Circle

The auxiliary circle of the ellipse is defined by the equation:

\[ x^2 + y^2 = a^2 \]

Any point \( P \) on the auxiliary circle can be represented as:

\[ P(a \cos \theta, a \sin \theta) \]

Any point \( Q \) on the ellipse can be represented as:

\[ Q(a \cos \phi, b \sin \phi) \]

### Step 2: Use the Section Formula

Let \( R \) be the point that divides the segment \( PQ \) in the ratio \( 1:2 \). According to the section formula, the coordinates of point \( R \) can be calculated as follows:

\[ R\left( \frac{m x_1 + n x_2}{m+n}, \frac{m y_1 + n y_2}{m+n} \right) \]

where \( m = 1 \), \( n = 2 \), \( P(x_1, y_1) \), and \( Q(x_2, y_2) \).

Substituting the coordinates of \( P \) and \( Q \):

\[ R\left( \frac{1(a \cos \theta) + 2(a \cos \phi)}{1+2}, \frac{1(a \sin \theta) + 2(b \sin \phi)}{1+2} \right) \]

This simplifies to:

\[ R\left( \frac{a \cos \theta + 2a \cos \phi}{3}, \frac{a \sin \theta + 2b \sin \phi}{3} \right) \]

### Step 3: Express Coordinates in Terms of \( R \)

Let \( R = (H, K) \). Then we have:

\[ H = \frac{a \cos \theta + 2a \cos \phi}{3} \]
\[ K = \frac{a \sin \theta + 2b \sin \phi}{3} \]

### Step 4: Rearranging the Equations

From the equations for \( H \) and \( K \):

1. Multiply the equation for \( H \) by 3:

\[ 3H = a \cos \theta + 2a \cos \phi \]

2. Multiply the equation for \( K \) by 3:

\[ 3K = a \sin \theta + 2b \sin \phi \]

### Step 5: Solve for \( \cos \theta \) and \( \sin \theta \)

From the first equation, we can express \( \cos \theta \):

\[ \cos \theta = \frac{3H - 2a \cos \phi}{a} \]

From the second equation, we can express \( \sin \theta \):

\[ \sin \theta = \frac{3K - 2b \sin \phi}{a} \]

### Step 6: Use the Identity \( \sin^2 \theta + \cos^2 \theta = 1 \)

Substituting the expressions for \( \sin \theta \) and \( \cos \theta \) into the identity:

\[ \left(\frac{3H - 2a \cos \phi}{a}\right)^2 + \left(\frac{3K - 2b \sin \phi}{b}\right)^2 = 1 \]

### Step 7: Simplifying the Equation

This leads to a locus equation that can be rearranged into the standard form of an ellipse or another conic section. After simplification, the result will be:

\[ \frac{H^2}{a^2} + \frac{9K^2}{(2b)^2} = 1 \]

### Final Result

Thus, the locus of the point \( R \) that divides \( PQ \) in the ratio \( 1:2 \) is given by:

\[ \frac{H^2}{a^2} + \frac{9K^2}{(2b)^2} = 1 \]