The equation of the locus of a point which moves so that its distance from the point (ak, 0) is k times its distance from the point `((a)/(k),0) (k ne 1)` is

Options

A. `x^(2)-y^(2)=a^(2)`
B. `2x^(2)-y^(2)=2a^(2)`
C. `xy=a^(2)`
D. `x^(2)+y^(2)=a^(2)`

Answer

To find the locus of a point \( P(x, y) \) that moves such that its distance from the point \( (ak, 0) \) is \( k \) times its distance from the point \( \left(\frac{a}{k}, 0\right) \) (where \( k \neq 1 \)), we can follow these steps:

### Step 1: Set up the distance equations
The distance from point \( P(x, y) \) to the point \( (ak, 0) \) is given by:
\[ d_1 = \sqrt{(x - ak)^2 + (y - 0)^2} = \sqrt{(x - ak)^2 + y^2} \]

The distance from point \( P(x, y) \) to the point \( \left(\frac{a}{k}, 0\right) \) is given by:
\[ d_2 = \sqrt{\left(x - \frac{a}{k}\right)^2 + (y - 0)^2} = \sqrt{\left(x - \frac{a}{k}\right)^2 + y^2} \]

According to the problem, we have:
\[ d_1 = k \cdot d_2 \]

### Step 2: Substitute the distance equations
Substituting the expressions for \( d_1 \) and \( d_2 \) into the equation gives:
\[ \sqrt{(x - ak)^2 + y^2} = k \cdot \sqrt{\left(x - \frac{a}{k}\right)^2 + y^2} \]

### Step 3: Square both sides
Squaring both sides to eliminate the square roots, we get:
\[ (x - ak)^2 + y^2 = k^2 \left(\left(x - \frac{a}{k}\right)^2 + y^2\right) \]

### Step 4: Expand both sides
Expanding both sides:
\[ (x^2 - 2akx + a^2k^2 + y^2) = k^2 \left(x^2 - \frac{2a}{k}x + \frac{a^2}{k^2} + y^2\right) \]

This simplifies to:
\[ x^2 - 2akx + a^2k^2 + y^2 = k^2x^2 - 2ax + a^2 + k^2y^2 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ x^2 - k^2x^2 + k^2y^2 - y^2 - 2akx + 2ax - a^2k^2 + a^2 = 0 \]

### Step 6: Factor out common terms
Factoring out the common terms leads to:
\[ (1 - k^2)x^2 + (k^2 - 1)y^2 - 2a(k - 1)x + (a^2 - a^2k^2) = 0 \]

### Step 7: Simplifying the equation
Since \( k^2 - 1 \neq 0 \) (as \( k \neq \pm 1 \)), we can divide the entire equation by \( k^2 - 1 \):
\[ x^2 + y^2 = a^2 \]

### Conclusion
Thus, the equation of the locus of point \( P \) is:
\[ x^2 + y^2 = a^2 \]
This represents a circle centered at the origin with radius \( a \).