Find the equation of the hyperbola whose foaci are `(0, 5) and (-2, 5)` and eccentricity 3.

Answer

To find the equation of the hyperbola whose foci are at (0, 5) and (-2, 5) with an eccentricity of 3, we can follow these steps:

### Step 1: Identify the coordinates of the foci
The foci are given as \( F_1(0, 5) \) and \( F_2(-2, 5) \).

### Step 2: Determine the center of the hyperbola
The center of the hyperbola is the midpoint of the line segment joining the foci. We can calculate it using the midpoint formula:
\[ \text{Center} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of the foci:
\[ \text{Center} = \left( \frac{0 + (-2)}{2}, \frac{5 + 5}{2} \right) = \left( -1, 5 \right) \]

### Step 3: Determine the distance between the foci
The distance \( 2c \) between the foci is given by:
\[ 2c = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of the foci:
\[ 2c = \sqrt{((-2) - 0)^2 + (5 - 5)^2} = \sqrt{(-2)^2} = 2 \]
Thus, \( c = 1 \).

### Step 4: Use the eccentricity to find \( a \)
The eccentricity \( e \) is given by the formula:
\[ e = \frac{c}{a} \]
Given that \( e = 3 \), we can rearrange this to find \( a \):
\[ 3 = \frac{1}{a} \implies a = \frac{1}{3} \]

### Step 5: Find \( b \) using the relationship \( c^2 = a^2 + b^2 \)
We know:
\[ c^2 = a^2 + b^2 \]
Substituting the values of \( c \) and \( a \):
\[ 1^2 = \left( \frac{1}{3} \right)^2 + b^2 \]
\[ 1 = \frac{1}{9} + b^2 \]
\[ b^2 = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9} \]

### Step 6: Write the equation of the hyperbola
The standard form of the equation of a hyperbola with a horizontal transverse axis centered at \( (h, k) \) is:
\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]
Substituting \( h = -1 \), \( k = 5 \), \( a^2 = \frac{1}{9} \), and \( b^2 = \frac{8}{9} \):
\[ \frac{(x + 1)^2}{\frac{1}{9}} - \frac{(y - 5)^2}{\frac{8}{9}} = 1 \]

### Step 7: Simplify the equation
Multiplying through by 9 to eliminate the denominators:
\[ 9 \cdot \frac{(x + 1)^2}{\frac{1}{9}} - 9 \cdot \frac{(y - 5)^2}{\frac{8}{9}} = 9 \]
This simplifies to:
\[ 9(x + 1)^2 - 8(y - 5)^2 = 72 \]

### Final Equation
Thus, the equation of the hyperbola is:
\[ 9(x + 1)^2 - 8(y - 5)^2 = 72 \]

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