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, we start by defining the points P and Q based on the given ellipse and its auxiliary circle.

### Step 1: Define the Points
The ellipse is given by the equation:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The auxiliary circle of the ellipse is given by:
\[ x^2 + y^2 = a^2 \]
Let \( P \) be a point on the auxiliary circle, which can be represented in parametric form as:
\[ P = (a \cos \theta, a \sin \theta) \]
Let \( Q \) be the corresponding point on the ellipse, represented as:
\[ Q = (a \cos \theta, b \sin \theta) \]

### Step 2: Find the Coordinates of the Point R
Let \( R(h, k) \) be the point that divides the segment \( PQ \) in the ratio \( 1:2 \). The coordinates of point \( R \) can be found using the section formula:
\[ R = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \]
where \( m = 1 \) and \( n = 2 \), \( P = (x_1, y_1) \) and \( Q = (x_2, y_2) \).

Substituting the coordinates of \( P \) and \( Q \):
\[ R = \left( \frac{1 \cdot (a \cos \theta) + 2 \cdot (a \cos \theta)}{1 + 2}, \frac{1 \cdot (b \sin \theta) + 2 \cdot (a \sin \theta)}{1 + 2} \right) \]
This simplifies to:
\[ R = \left( \frac{3a \cos \theta}{3}, \frac{b \sin \theta + 2a \sin \theta}{3} \right) \]
Thus,
\[ R = (a \cos \theta, \frac{(b + 2a) \sin \theta}{3}) \]

### Step 3: Express R in Terms of h and k
From the coordinates of \( R \):
\[ h = a \cos \theta \]
\[ k = \frac{(b + 2a) \sin \theta}{3} \]

### Step 4: Eliminate θ
To find the locus, we need to eliminate \( \theta \). From \( h = a \cos \theta \), we can express \( \cos \theta \):
\[ \cos \theta = \frac{h}{a} \]
From \( k = \frac{(b + 2a) \sin \theta}{3} \), we can express \( \sin \theta \):
\[ \sin \theta = \frac{3k}{b + 2a} \]

Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ \left(\frac{h}{a}\right)^2 + \left(\frac{3k}{b + 2a}\right)^2 = 1 \]

### Step 5: Simplify the Equation
Squaring both terms:
\[ \frac{h^2}{a^2} + \frac{9k^2}{(b + 2a)^2} = 1 \]

### Step 6: Rearranging the Equation
Multiplying through by \( a^2(b + 2a)^2 \):
\[ h^2(b + 2a)^2 + 9k^2a^2 = a^2(b + 2a)^2 \]

This gives us the equation of the locus of point \( R \):
\[ \frac{h^2}{a^2} + \frac{9k^2}{(b + 2a)^2} = 1 \]

### Final Locus Equation
Thus, the locus of the point that divides \( PQ \) in the ratio \( 1:2 \) is:
\[ \frac{x^2}{a^2} + \frac{9y^2}{(b + 2a)^2} = 1 \]