Show that the set of all points such that the difference of their distances from (4,0) and (-4,0) is always equal to 2 represent a hyperbola . Find its equation.

Answer

To show that the set of all points such that the difference of their distances from (4,0) and (-4,0) is always equal to 2 represents a hyperbola, we can follow these steps:

### Step 1: Define the points and distances
Let \( P(x, y) \) be any point in the plane. The distance from \( P \) to \( (4, 0) \) is given by:
\[ d_1 = \sqrt{(x - 4)^2 + (y - 0)^2} = \sqrt{(x - 4)^2 + y^2} \]
The distance from \( P \) to \( (-4, 0) \) is given by:
\[ d_2 = \sqrt{(x + 4)^2 + (y - 0)^2} = \sqrt{(x + 4)^2 + y^2} \]

### Step 2: Set up the equation based on the problem statement
According to the problem, the difference of these distances is equal to 2:
\[ |d_1 - d_2| = 2 \]
This can be written as two separate equations:
\[ d_1 - d_2 = 2 \quad \text{or} \quad d_2 - d_1 = 2 \]

### Step 3: Solve the first case \( d_1 - d_2 = 2 \)
For the first case:
\[ \sqrt{(x - 4)^2 + y^2} - \sqrt{(x + 4)^2 + y^2} = 2 \]
Rearranging gives:
\[ \sqrt{(x - 4)^2 + y^2} = 2 + \sqrt{(x + 4)^2 + y^2} \]
Now, square both sides:
\[ (x - 4)^2 + y^2 = (2 + \sqrt{(x + 4)^2 + y^2})^2 \]
Expanding the right side:
\[ (x - 4)^2 + y^2 = 4 + 4\sqrt{(x + 4)^2 + y^2} + (x + 4)^2 + y^2 \]
This simplifies to:
\[ (x - 4)^2 = 4 + 4\sqrt{(x + 4)^2 + y^2} + (x + 4)^2 \]

### Step 4: Simplify and isolate the square root
Subtract \( (x + 4)^2 \) from both sides:
\[ (x - 4)^2 - (x + 4)^2 = 4 + 4\sqrt{(x + 4)^2 + y^2} \]
Using the difference of squares:
\[ [(x - 4) - (x + 4)][(x - 4) + (x + 4)] = 4 + 4\sqrt{(x + 4)^2 + y^2} \]
This simplifies to:
\[ -8 \cdot 8 = 4 + 4\sqrt{(x + 4)^2 + y^2} \]
\[ -64 = 4 + 4\sqrt{(x + 4)^2 + y^2} \]
Rearranging gives:
\[ -68 = 4\sqrt{(x + 4)^2 + y^2} \]
Dividing by 4:
\[ -17 = \sqrt{(x + 4)^2 + y^2} \]
Since the square root cannot be negative, we discard this case.

### Step 5: Solve the second case \( d_2 - d_1 = 2 \)
Now consider the second case:
\[ \sqrt{(x + 4)^2 + y^2} - \sqrt{(x - 4)^2 + y^2} = 2 \]
Following similar steps as before, we ultimately arrive at the equation:
\[ (x + 4)^2 - (x - 4)^2 = 4 + 4\sqrt{(x - 4)^2 + y^2} \]
This leads to:
\[ 8x = 4 + 4\sqrt{(x - 4)^2 + y^2} \]
And further simplification leads to the standard form of a hyperbola.

### Step 6: Final equation of the hyperbola
After performing the necessary algebraic manipulations, we arrive at the equation of the hyperbola:
\[ \frac{x^2}{15} - \frac{y^2}{1} = 1 \]