A and A' be the points `(5,0)` and `(-5,0)` respectively. Find the equation of the set of all points `P(x,y)` such that `|AP - A'P|=6`

Answer

To solve the problem, we need to find the equation of the set of all points \( P(x, y) \) such that \( |AP - A'P| = 6 \), where \( A(5, 0) \) and \( A'(-5, 0) \).

### Step-by-Step Solution:

1. **Identify Points and Distances**:
- Let \( A = (5, 0) \) and \( A' = (-5, 0) \).
- Let \( P = (x, y) \).
- We need to express the distances \( AP \) and \( A'P \).

2. **Distance Formulas**:
- The distance \( AP \) is given by:
\[ AP = \sqrt{(x - 5)^2 + (y - 0)^2} = \sqrt{(x - 5)^2 + y^2} \]
- The distance \( A'P \) is given by:
\[ A'P = \sqrt{(x + 5)^2 + (y - 0)^2} = \sqrt{(x + 5)^2 + y^2} \]

3. **Set Up the Equation**:
- According to the problem, we have:
\[ |AP - A'P| = 6 \]
- This leads to two cases:
1. \( AP - A'P = 6 \)
2. \( A'P - AP = 6 \)

4. **Case 1: \( AP - A'P = 6 \)**:
- Squaring both sides:
\[ \left( \sqrt{(x - 5)^2 + y^2} - \sqrt{(x + 5)^2 + y^2} \right)^2 = 36 \]
- Expanding:
\[ (x - 5)^2 + y^2 - 2\sqrt{((x - 5)^2 + y^2)((x + 5)^2 + y^2)} + (x + 5)^2 + y^2 = 36 \]
- Combine like terms:
\[ 2x^2 + 50 - 2\sqrt{((x - 5)^2 + y^2)((x + 5)^2 + y^2)} = 36 \]
- Rearranging gives:
\[ 2\sqrt{((x - 5)^2 + y^2)((x + 5)^2 + y^2)} = 2x^2 + 14 \]
- Dividing by 2:
\[ \sqrt{((x - 5)^2 + y^2)((x + 5)^2 + y^2)} = x^2 + 7 \]
- Squaring again:
\[ ((x - 5)^2 + y^2)((x + 5)^2 + y^2) = (x^2 + 7)^2 \]

5. **Case 2: \( A'P - AP = 6 \)**:
- Following a similar process as above, we will arrive at another equation.

6. **Combine and Simplify**:
- After solving both cases, we will find that the resulting equations represent a hyperbola.

7. **Final Equation**:
- The final equation of the set of all points \( P(x, y) \) such that \( |AP - A'P| = 6 \) is:
\[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \]
- This represents a hyperbola centered at the origin.