Find the equation of the hyperbola whose foci are `(4,1), (8,1)` and whose eccentricity is 2.

Answer

To find the equation of the hyperbola with foci at (4, 1) and (8, 1) and an eccentricity of 2, we can follow these steps:

### Step 1: Identify the center of the hyperbola
The center of the hyperbola is the midpoint of the foci.

\[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{4 + 8}{2}, \frac{1 + 1}{2} \right) = \left( \frac{12}{2}, \frac{2}{2} \right) = (6, 1) \]

### Step 2: Determine the distance between the foci
The distance between the foci is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(8 - 4)^2 + (1 - 1)^2} = \sqrt{4^2} = 4 \]

### Step 3: Calculate \(2c\) and find \(c\)
Since the distance between the foci is \(2c\):

\[ 2c = 4 \implies c = 2 \]

### Step 4: Relate \(a\), \(b\), and \(c\) using eccentricity
The eccentricity \(e\) is given as 2. The relationship between \(a\), \(b\), and \(c\) for hyperbolas is:

\[ e = \frac{c}{a} \implies 2 = \frac{2}{a} \implies a = 1 \]

### Step 5: Calculate \(b^2\)
Using the relationship \(c^2 = a^2 + b^2\):

\[ c^2 = 2^2 = 4, \quad a^2 = 1^2 = 1 \]
\[ 4 = 1 + b^2 \implies b^2 = 4 - 1 = 3 \]

### Step 6: Write the equation of the hyperbola
The standard form of the equation of a hyperbola centered at \((h, k)\) is:

\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]

Substituting \(h = 6\), \(k = 1\), \(a^2 = 1\), and \(b^2 = 3\):

\[ \frac{(x - 6)^2}{1} - \frac{(y - 1)^2}{3} = 1 \]

### Step 7: Expand the equation
We can expand this equation:

\[ (x - 6)^2 - \frac{(y - 1)^2}{3} = 1 \]

Multiplying through by 3 to eliminate the fraction:

\[ 3((x - 6)^2) - (y - 1)^2 = 3 \]

### Final Equation
Thus, the equation of the hyperbola is:

\[ 3(x - 6)^2 - (y - 1)^2 - 3 = 0 \]