The equations of the latus recta of the hyperbola ` 9x^(2) -16y^(2) -18x -32y -151 =0` are

Options

A. x=6, x+4=0
B. x=-6, x-14=0
C. x=3, x+4=0
D. x=2, x-11=0

Answer

To find the equations of the latus recta of the hyperbola given by the equation \( 9x^2 - 16y^2 - 18x - 32y - 151 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the given equation:
\[ 9x^2 - 16y^2 - 18x - 32y - 151 = 0 \]

First, we will rearrange the equation by grouping the \(x\) and \(y\) terms:
\[ 9x^2 - 18x - 16y^2 - 32y = 151 \]

### Step 2: Complete the square for \(x\) and \(y\)
For the \(x\) terms:
\[ 9(x^2 - 2x) \]
To complete the square, we take half of \(-2\) (which is \(-1\)), square it (getting \(1\)), and adjust:
\[ 9(x^2 - 2x + 1 - 1) = 9((x - 1)^2 - 1) = 9(x - 1)^2 - 9 \]

For the \(y\) terms:
\[ -16(y^2 + 2y) \]
Completing the square here, we take half of \(2\) (which is \(1\)), square it (getting \(1\)), and adjust:
\[ -16(y^2 + 2y + 1 - 1) = -16((y + 1)^2 - 1) = -16(y + 1)^2 + 16 \]

### Step 3: Substitute back into the equation
Substituting back into the equation gives:
\[ 9(x - 1)^2 - 9 - 16(y + 1)^2 + 16 = 151 \]
This simplifies to:
\[ 9(x - 1)^2 - 16(y + 1)^2 + 7 = 151 \]
So,
\[ 9(x - 1)^2 - 16(y + 1)^2 = 144 \]

### Step 4: Divide by 144 to get the standard form
Dividing the entire equation by \(144\):
\[ \frac{9(x - 1)^2}{144} - \frac{16(y + 1)^2}{144} = 1 \]
This simplifies to:
\[ \frac{(x - 1)^2}{16} - \frac{(y + 1)^2}{9} = 1 \]

### Step 5: Identify parameters \(a\) and \(b\)
From the standard form, we can identify:
- \(a^2 = 16 \Rightarrow a = 4\)
- \(b^2 = 9 \Rightarrow b = 3\)

### Step 6: Find the eccentricity \(e\)
The eccentricity \(e\) of the hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]

### Step 7: Find the equations of the latus recta
The equations of the latus recta are given by:
\[ x = x_1 \pm ae \]
Where \(x_1\) is the x-coordinate of the center of the hyperbola, which is \(1\):
\[ x = 1 \pm (4 \cdot \frac{5}{4}) = 1 \pm 5 \]

Thus, we have:
1. \(x = 6\)
2. \(x = -4\)

### Final Answer
The equations of the latus recta are:
\[ x = 6 \quad \text{and} \quad x = -4 \]