Equation of the hyperbola of eccentricity 3 and the distance between whose foci is 24 is

Options

A. `x^(2)-8y^(2)=128`
B. `x^(2)-16y^(2)=128`
C. `16x^(2)-y^(2)=128`
D. `8x^(2)-y^(2)=128`

Answer

To find the equation of the hyperbola with an eccentricity of 3 and a distance between its foci of 24, we can follow these steps:

### Step 1: Understand the relationship between the foci, eccentricity, and semi-major axis
The distance between the foci of a hyperbola is given by the formula:
\[ 2c = 24 \]
where \( c \) is the distance from the center to each focus. Therefore, we can find \( c \):
\[ c = \frac{24}{2} = 12 \]

### Step 2: Use the eccentricity to find \( a \)
The eccentricity \( e \) of a hyperbola is defined as:
\[ e = \frac{c}{a} \]
Given that \( e = 3 \), we can substitute the value of \( c \):
\[ 3 = \frac{12}{a} \]
Now, we can solve for \( a \):
\[ a = \frac{12}{3} = 4 \]

### Step 3: Calculate \( a^2 \)
Now that we have \( a \), we can find \( a^2 \):
\[ a^2 = 4^2 = 16 \]

### Step 4: Find \( b^2 \) using the relationship between \( a \), \( b \), and \( c \)
We know the relationship:
\[ c^2 = a^2 + b^2 \]
Substituting the known values:
\[ 12^2 = 16 + b^2 \]
\[ 144 = 16 + b^2 \]
Now, solve for \( b^2 \):
\[ b^2 = 144 - 16 = 128 \]

### Step 5: Write the equation of the hyperbola
The standard form of the equation of a hyperbola centered at the origin is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Substituting the values of \( a^2 \) and \( b^2 \):
\[ \frac{x^2}{16} - \frac{y^2}{128} = 1 \]

### Step 6: Simplify the equation
To simplify, we can multiply the entire equation by 128 to eliminate the denominators:
\[ 8x^2 - y^2 = 128 \]

### Final Answer
Thus, the equation of the hyperbola is:
\[ 8x^2 - y^2 = 128 \]

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