The locus of a point P which divides the line joining `(1,0) and (2 costheta, 2 sintheta)` intermally in the ratio `1:2` for all `theta`. is

Options

A. straight line
B. circle
C. pair of straight line
D. parabola

Answer

To find the locus of the point \( P \) that divides the line segment joining the points \( (1, 0) \) and \( (2 \cos \theta, 2 \sin \theta) \) internally in the ratio \( 1:2 \), we can follow these steps:

### Step 1: Identify the coordinates of the points
Let:
- Point \( A = (1, 0) \)
- Point \( B = (2 \cos \theta, 2 \sin \theta) \)

### Step 2: Use the section formula
The coordinates of point \( P \) that divides the line segment \( AB \) in the ratio \( m:n = 1:2 \) can be found using the section formula:
\[ P(x, y) = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \]
where \( (x_1, y_1) = (1, 0) \) and \( (x_2, y_2) = (2 \cos \theta, 2 \sin \theta) \).

### Step 3: Substitute the values into the formula
Substituting \( m = 1 \), \( n = 2 \), \( x_1 = 1 \), \( y_1 = 0 \), \( x_2 = 2 \cos \theta \), and \( y_2 = 2 \sin \theta \):
\[ x = \frac{1 \cdot (2 \cos \theta) + 2 \cdot (1)}{1 + 2} = \frac{2 \cos \theta + 2}{3} = \frac{2(1 + \cos \theta)}{3} \]
\[ y = \frac{1 \cdot (2 \sin \theta) + 2 \cdot (0)}{1 + 2} = \frac{2 \sin \theta}{3} \]

### Step 4: Express \( \cos \theta \) and \( \sin \theta \) in terms of \( x \) and \( y \)
From the equations derived:
1. \( x = \frac{2(1 + \cos \theta)}{3} \) implies \( 1 + \cos \theta = \frac{3x}{2} \) or \( \cos \theta = \frac{3x}{2} - 1 \).
2. \( y = \frac{2 \sin \theta}{3} \) implies \( \sin \theta = \frac{3y}{2} \).

### Step 5: Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \)
Substituting the expressions for \( \sin \theta \) and \( \cos \theta \):
\[ \left(\frac{3y}{2}\right)^2 + \left(\frac{3x}{2} - 1\right)^2 = 1 \]

### Step 6: Simplify the equation
Expanding both terms:
\[ \frac{9y^2}{4} + \left(\frac{3x}{2} - 1\right)^2 = 1 \]
\[ \frac{9y^2}{4} + \left(\frac{9x^2}{4} - 3x + 1\right) = 1 \]
Combining the terms:
\[ \frac{9y^2}{4} + \frac{9x^2}{4} - 3x + 1 = 1 \]
Subtracting 1 from both sides:
\[ \frac{9y^2}{4} + \frac{9x^2}{4} - 3x = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 9y^2 + 9x^2 - 12x = 0 \]
Rearranging gives:
\[ 9x^2 + 9y^2 - 12x = 0 \]

### Step 7: Complete the square
Rearranging:
\[ 9(x^2 - \frac{12}{9}x + y^2) = 0 \]
Completing the square for \( x \):
\[ 9\left((x - \frac{2}{3})^2 + y^2 - \frac{4}{9}\right) = 0 \]
Thus:
\[ (x - \frac{2}{3})^2 + y^2 = \frac{4}{9} \]

### Conclusion
This is the equation of a circle with center \( \left(\frac{2}{3}, 0\right) \) and radius \( \frac{2}{3} \).

### Final Answer
The locus of the point \( P \) is a circle given by:
\[ \left(x - \frac{2}{3}\right)^2 + y^2 = \left(\frac{2}{3}\right)^2 \]