A (c,0) and B (-c,0) are two points . If P is a point such that PA + PB = 2a where `0 lt c lt a,` then find the locus of P.

Answer

To find the locus of the point P such that PA + PB = 2a, where A(c, 0) and B(-c, 0) are two fixed points and \(0 < c < a\), we can follow these steps:

### Step 1: Define the Points and the Point P
Let the coordinates of point P be (x, y). The coordinates of points A and B are given as A(c, 0) and B(-c, 0).

### Step 2: Use the Distance Formula
Using the distance formula, we can express PA and PB as follows:
- The distance PA from point P to A is given by:
\[ PA = \sqrt{(x - c)^2 + (y - 0)^2} = \sqrt{(x - c)^2 + y^2} \]
- The distance PB from point P to B is given by:
\[ PB = \sqrt{(x + c)^2 + (y - 0)^2} = \sqrt{(x + c)^2 + y^2} \]

### Step 3: Set Up the Given Condition
According to the problem, we have:
\[ PA + PB = 2a \]

### Step 4: Square Both Sides
To eliminate the square roots, we will first square both sides of the equation. However, it is more convenient to manipulate the equation first. We can use the identity:
\[ PA^2 - PB^2 = (PA + PB)(PA - PB) \]
This means we will find PA^2 and PB^2.

### Step 5: Calculate PA^2 and PB^2
Calculating PA^2:
\[ PA^2 = (x - c)^2 + y^2 = x^2 - 2xc + c^2 + y^2 \]
Calculating PB^2:
\[ PB^2 = (x + c)^2 + y^2 = x^2 + 2xc + c^2 + y^2 \]

### Step 6: Find PA^2 - PB^2
Now we can find:
\[ PA^2 - PB^2 = (x^2 - 2xc + c^2 + y^2) - (x^2 + 2xc + c^2 + y^2) \]
This simplifies to:
\[ PA^2 - PB^2 = -4xc \]

### Step 7: Substitute into the Identity
Using the identity:
\[ PA^2 - PB^2 = (PA + PB)(PA - PB) \]
Substituting \(PA + PB = 2a\):
\[ -4xc = 2a(PA - PB) \]

### Step 8: Find PA - PB
From the above equation, we can express \(PA - PB\):
\[ PA - PB = \frac{-4xc}{2a} = \frac{-2xc}{a} \]

### Step 9: Add and Solve for PA
Now, we can add the equations:
\[ PA + PB = 2a \quad \text{and} \quad PA - PB = \frac{-2xc}{a} \]
Adding these two equations:
\[ 2PA = 2a + \frac{-2xc}{a} \]
Thus,
\[ PA = a - \frac{xc}{a} \]

### Step 10: Square PA
Now, square PA:
\[ PA^2 = \left(a - \frac{xc}{a}\right)^2 \]
Expanding this gives:
\[ PA^2 = a^2 - 2a\left(\frac{xc}{a}\right) + \left(\frac{xc}{a}\right)^2 = a^2 - 2xc + \frac{x^2c^2}{a^2} \]

### Step 11: Set PA^2 Equal to the Expression
We know from the distance formula:
\[ PA^2 = (x - c)^2 + y^2 \]
Setting these equal gives us:
\[ (x - c)^2 + y^2 = a^2 - 2xc + \frac{x^2c^2}{a^2} \]

### Step 12: Rearranging the Equation
After simplifying and rearranging, we will arrive at the equation of a circle:
\[ \frac{x^2}{a^2 - c^2} + \frac{y^2}{a^2} = 1 \]

### Conclusion
Thus, the locus of point P is an ellipse given by:
\[ \frac{x^2}{a^2 - c^2} + \frac{y^2}{a^2} = 1 \]