The length of the latus rectum of the hyperbola ` 9x^(2) -16y^(2) +72x -32y- 16 =0 ` is

Options

A. `9//2`
B. `32//3`
C. `11//5`
D. `21//5`

Answer

To find the length of the latus rectum of the hyperbola given by the equation \( 9x^2 - 16y^2 + 72x - 32y - 16 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation:
\[ 9x^2 - 16y^2 + 72x - 32y - 16 = 0 \]

Rearranging gives us:
\[ 9x^2 + 72x - 16y^2 - 32y = 16 \]

### Step 2: Complete the square for \(x\) and \(y\)
**For \(x\):**
Factor out the coefficient of \(x^2\):
\[ 9(x^2 + 8x) - 16y^2 - 32y = 16 \]
Complete the square for \(x\):
\[ x^2 + 8x = (x + 4)^2 - 16 \]
Thus, we have:
\[ 9((x + 4)^2 - 16) - 16y^2 - 32y = 16 \]
This simplifies to:
\[ 9(x + 4)^2 - 144 - 16y^2 - 32y = 16 \]

**For \(y\):**
Factor out the coefficient of \(y^2\):
\[ -16(y^2 + 2y) = -16((y + 1)^2 - 1) = -16(y + 1)^2 + 16 \]
Substituting back, we get:
\[ 9(x + 4)^2 - 144 - 16(y + 1)^2 + 16 = 16 \]
Combining the constants gives:
\[ 9(x + 4)^2 - 16(y + 1)^2 - 112 = 0 \]

### Step 3: Rearranging to standard hyperbola form
Now, we can rearrange:
\[ 9(x + 4)^2 - 16(y + 1)^2 = 112 \]
Dividing through by 112 gives:
\[ \frac{(x + 4)^2}{\frac{112}{9}} - \frac{(y + 1)^2}{\frac{112}{16}} = 1 \]

### Step 4: Identify \(a^2\) and \(b^2\)
From the equation, we identify:
\[ a^2 = \frac{112}{9}, \quad b^2 = \frac{112}{16} \]

### Step 5: Calculate the length of the latus rectum
The formula for the length of the latus rectum \(L\) of a hyperbola is given by:
\[ L = \frac{2b^2}{a} \]
First, we need to find \(a\):
\[ a = \sqrt{\frac{112}{9}} = \frac{4\sqrt{7}}{3} \]
Now calculate \(b^2\):
\[ b^2 = \frac{112}{16} = 7 \]
Thus:
\[ L = \frac{2 \cdot 7}{\frac{4\sqrt{7}}{3}} = \frac{14 \cdot 3}{4\sqrt{7}} = \frac{42}{4\sqrt{7}} = \frac{21}{2\sqrt{7}} = \frac{21\sqrt{7}}{14} = \frac{3\sqrt{7}}{2} \]

### Final Answer
The length of the latus rectum of the hyperbola is:
\[ \frac{21}{2\sqrt{7}} \text{ or } \frac{3\sqrt{7}}{2} \]