A(a,0) , B(-a,0) are two point . If a point P moves such that `anglePAB - anglePBA = 2alpha` then find the locus of P .

Answer

To find the locus of point P such that the condition \( \angle PAB - \angle PBA = 2\alpha \) holds, we can follow these steps:

### Step 1: Define Points and Angles
Let the points be defined as follows:
- Point A: \( A(a, 0) \)
- Point B: \( B(-a, 0) \)
- Point P: \( P(x, y) \)

### Step 2: Identify Angles
Let:
- \( \theta_1 = \angle PAB \)
- \( \theta_2 = \angle PBA \)

According to the problem, we have:
\[ \theta_1 - \theta_2 = 2\alpha \]

### Step 3: Calculate Tangents of Angles
Using the definition of tangent in terms of the coordinates, we can express:
- \( \tan(\theta_1) = \frac{y}{a - x} \)
- \( \tan(\theta_2) = \frac{y}{x + a} \)

### Step 4: Apply the Tangent Difference Formula
We can use the tangent difference formula:
\[ \tan(\theta_1 - \theta_2) = \frac{\tan(\theta_1) - \tan(\theta_2)}{1 + \tan(\theta_1) \tan(\theta_2)} \]
Substituting the values we calculated:
\[ \tan(2\alpha) = \frac{\frac{y}{a - x} - \frac{y}{x + a}}{1 + \frac{y}{a - x} \cdot \frac{y}{x + a}} \]

### Step 5: Simplify the Expression
Now, we simplify the left-hand side:
\[ \tan(2\alpha) = \frac{y \left( \frac{1}{a - x} - \frac{1}{x + a} \right)}{1 + \frac{y^2}{(a - x)(x + a)}} \]
The numerator simplifies to:
\[ \frac{y \left( (x + a) - (a - x) \right)}{(a - x)(x + a)} = \frac{y(2x)}{(a - x)(x + a)} \]

### Step 6: Set Up the Equation
Now we equate this to \( \tan(2\alpha) \):
\[ \frac{y(2x)}{(a - x)(x + a)} = \tan(2\alpha) \]

### Step 7: Rearranging Terms
Cross-multiplying gives:
\[ 2xy = \tan(2\alpha) \cdot ((a - x)(x + a)) \]
Expanding the right-hand side:
\[ 2xy = \tan(2\alpha) \cdot (a^2 - x^2) \]

### Step 8: Final Locus Equation
Rearranging the equation leads to:
\[ 2xy - \tan(2\alpha) x^2 + \tan(2\alpha) a^2 = 0 \]
This can be rewritten as:
\[ x^2 \tan(2\alpha) + 2xy = a^2 \tan(2\alpha) \]

### Final Answer
Thus, the locus of point P is given by:
\[ x^2 - a^2 \tan(2\alpha) + 2xy = 0 \]