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

Answer

To find the locus of point P such that the difference between angles \( \angle PAB \) and \( \angle PBA \) is \( \frac{\pi}{2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Points A and B**:
Let point A be \( A(a, 0) \) and point B be \( B(-a, 0) \).

2. **Define Point P**:
Let point P have coordinates \( P(x, y) \).

3. **Calculate Angles**:
We need to express \( \tan(\angle PAB) \) and \( \tan(\angle PBA) \):
- For \( \angle PAB \):
\[ \tan(\angle PAB) = \frac{y}{x - a} \]
- For \( \angle PBA \):
\[ \tan(\angle PBA) = \frac{y}{x + a} \]

4. **Set Up the Equation**:
According to the problem, we have:
\[ \angle PAB - \angle PBA = \frac{\pi}{2} \]
This implies:
\[ \tan(\angle PAB - \angle PBA) = \tan\left(\frac{\pi}{2}\right) \]
Since \( \tan\left(\frac{\pi}{2}\right) \) is undefined, we can use the formula for the tangent of the difference of two angles:
\[ \tan(\angle PAB - \angle PBA) = \frac{\tan(\angle PAB) - \tan(\angle PBA)}{1 + \tan(\angle PAB) \tan(\angle PBA)} \]
Setting this equal to infinity means the denominator must be zero:
\[ 1 + \tan(\angle PAB) \tan(\angle PBA) = 0 \]

5. **Substitute the Tangents**:
Substituting the expressions for \( \tan(\angle PAB) \) and \( \tan(\angle PBA) \):
\[ 1 + \left(\frac{y}{x - a}\right)\left(\frac{y}{x + a}\right) = 0 \]

6. **Simplify the Equation**:
This leads to:
\[ 1 + \frac{y^2}{(x - a)(x + a)} = 0 \]
Rearranging gives:
\[ \frac{y^2}{x^2 - a^2} = -1 \]
Multiplying through by \( (x^2 - a^2) \) gives:
\[ y^2 = - (x^2 - a^2) \]
This simplifies to:
\[ x^2 + y^2 = a^2 \]

7. **Conclusion**:
The locus of point P is a circle with center at the origin (0,0) and radius \( a \).

### Final Result:
The locus of point P is given by the equation:
\[ x^2 + y^2 = a^2 \]