The locus of the point which moves so that the difference of its distance from the points (5, 0) and (-5, 0) is 2, is ……………

Answer

To find the locus of the point that moves such that the difference of its distances from the points (5, 0) and (-5, 0) is 2, we can follow these steps:

### Step 1: Understand the Definition of a Hyperbola
The locus of points where the difference of distances from two fixed points is constant defines a hyperbola. In this case, the fixed points are (5, 0) and (-5, 0).

### Step 2: Set Up the Distance Equation
Let the point be (x, y). The distance from (x, y) to (5, 0) is given by:
\[ d_1 = \sqrt{(x - 5)^2 + (y - 0)^2} = \sqrt{(x - 5)^2 + y^2} \]

The distance from (x, y) to (-5, 0) is given by:
\[ d_2 = \sqrt{(x + 5)^2 + (y - 0)^2} = \sqrt{(x + 5)^2 + y^2} \]

### Step 3: Set Up the Difference of Distances
According to the problem, the difference of these distances is 2:
\[ |d_1 - d_2| = 2 \]

This gives us two cases to consider:
1. \( d_1 - d_2 = 2 \)
2. \( d_2 - d_1 = 2 \)

### Step 4: Solve the First Case
For the first case:
\[ \sqrt{(x - 5)^2 + y^2} - \sqrt{(x + 5)^2 + y^2} = 2 \]

Squaring both sides:
\[ (x - 5)^2 + y^2 - 2\sqrt{(x - 5)^2 + y^2}\sqrt{(x + 5)^2 + y^2} + (x + 5)^2 + y^2 = 4 \]

### Step 5: Simplify the Equation
Combine like terms:
\[ (x - 5)^2 + (x + 5)^2 + 2y^2 - 4 = 2\sqrt{(x - 5)^2 + y^2}\sqrt{(x + 5)^2 + y^2} \]

### Step 6: Solve the Second Case
For the second case:
\[ \sqrt{(x + 5)^2 + y^2} - \sqrt{(x - 5)^2 + y^2} = 2 \]

Following similar steps as above, we would arrive at a similar equation.

### Step 7: Final Form of the Hyperbola
After solving both cases, we can conclude that the locus of the point is a hyperbola centered at the origin with foci at (5, 0) and (-5, 0) and the equation of the hyperbola can be expressed as:
\[ \frac{x^2}{1} - \frac{y^2}{21} = 1 \]

### Conclusion
Thus, the locus of the point is a hyperbola given by the equation:
\[ \frac{x^2}{1} - \frac{y^2}{21} = 1 \]