The equation of the hyperbola whose foci are `(pm 4, 0)` and length of latus rectum is 12 is

Options

A. A. `(x^(2))/( 12) - (y^(2))/(4) = 1 `
B. B. `(y^(2))/(4)-(x^(2))/(12)=1`
C. C. `(y^(2))/(12)-(x^(2))/(4)=1`
D. D. `(x^(2))/(4)-(y^(2))/(12)=1`

Answer

To find the equation of the hyperbola whose foci are at (±4, 0) and whose length of the latus rectum is 12, we can follow these steps:

### Step 1: Identify the values of \( c \) and \( a \)
The foci of the hyperbola are given as \( (±4, 0) \). In the standard form of a hyperbola centered at the origin, the foci are located at \( (±c, 0) \). Thus, we have:
\[ c = 4 \]

### Step 2: Use the relationship between \( c \), \( a \), and \( b \)
For hyperbolas, the relationship between \( a \), \( b \), and \( c \) is given by:
\[ c^2 = a^2 + b^2 \]
We also know that the length of the latus rectum \( L \) is given by:
\[ L = \frac{2b^2}{a} \]
Given that the length of the latus rectum is 12, we can set up the equation:
\[ \frac{2b^2}{a} = 12 \]
From this, we can express \( b^2 \) in terms of \( a \):
\[ b^2 = 6a \]

### Step 3: Substitute \( b^2 \) into the equation for \( c^2 \)
Now substituting \( b^2 = 6a \) into the equation \( c^2 = a^2 + b^2 \):
\[ 16 = a^2 + 6a \]

### Step 4: Rearrange and solve for \( a \)
Rearranging gives us a quadratic equation:
\[ a^2 + 6a - 16 = 0 \]
To solve this quadratic equation, we can factor it:
\[ (a + 8)(a - 2) = 0 \]
Thus, the solutions for \( a \) are:
\[ a = 2 \quad \text{(valid, since distance cannot be negative)} \]
\[ a = -8 \quad \text{(not valid)} \]

### Step 5: Calculate \( b^2 \)
Now that we have \( a = 2 \), we can find \( b^2 \):
\[ b^2 = 6a = 6 \times 2 = 12 \]

### Step 6: Write the equation of the hyperbola
The standard form of the equation of a hyperbola that opens horizontally is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Substituting \( a^2 = 4 \) and \( b^2 = 12 \):
\[ \frac{x^2}{4} - \frac{y^2}{12} = 1 \]

### Final Answer
Thus, the equation of the hyperbola is:
\[ \frac{x^2}{4} - \frac{y^2}{12} = 1 \]