The coordinates of the point `A` and `B` are `(a,0)` and `(-a ,0),` respectively. If a point `P` moves so that `P A^2-P B^2=2k^2,` when `k` is constant, then find the equation to the locus of the point `P`.

Explanation

To find the equation of the locus of the point \( P \) given the coordinates of points \( A \) and \( B \) and the condition \( PA^2 - PB^2 = 2k^2 \), we can follow these steps:

### Step 1: Define the coordinates of points A, B, and P
Let the coordinates of point \( A \) be \( (a, 0) \) and the coordinates of point \( B \) be \( (-a, 0) \). Denote the coordinates of point \( P \) as \( (x, y) \).

### Step 2: Calculate the distances \( PA \) and \( PB \)
The distance \( PA \) from point \( P \) to point \( A \) is given by:
\[ PA = \sqrt{(x - a)^2 + (y - 0)^2} = \sqrt{(x - a)^2 + y^2} \]
The distance \( PB \) from point \( P \) to point \( B \) is given by:
\[ PB = \sqrt{(x + a)^2 + (y - 0)^2} = \sqrt{(x + a)^2 + y^2} \]

### Step 3: Square the distances
Now, we square both distances:
\[ PA^2 = (x - a)^2 + y^2 \]
\[ PB^2 = (x + a)^2 + y^2 \]

### Step 4: Set up the equation based on the given condition
According to the problem, we have:
\[ PA^2 - PB^2 = 2k^2 \]
Substituting the squared distances:
\[ ((x - a)^2 + y^2) - ((x + a)^2 + y^2) = 2k^2 \]

### Step 5: Simplify the equation
The \( y^2 \) terms cancel out:
\[ (x - a)^2 - (x + a)^2 = 2k^2 \]
Expanding both squares:
\[ (x^2 - 2ax + a^2) - (x^2 + 2ax + a^2) = 2k^2 \]
This simplifies to:
\[ -4ax = 2k^2 \]

### Step 6: Rearranging the equation
Dividing both sides by 2:
\[ -2ax = k^2 \]
Rearranging gives:
\[ 2ax + k^2 = 0 \]

### Step 7: Final form of the locus equation
Thus, the equation of the locus of point \( P \) is:
\[ 2ax + k^2 = 0 \]