The diameter of `16x^(2)-9y^(2)=144` which is conjugate to `x=2y` is

Options

A. `y=(16)/(9)x`
B. `y=(32)/(9)x`
C. `x=(16)/(9)y`
D. `x=(32)/(9)y`

Answer

To find the diameter of the hyperbola given by the equation \(16x^2 - 9y^2 = 144\) that is conjugate to the line \(x = 2y\), we can follow these steps:

### Step 1: Rewrite the equation of the hyperbola in standard form
We start with the equation:
\[ 16x^2 - 9y^2 = 144 \]
Dividing both sides by 144 gives:
\[ \frac{16x^2}{144} - \frac{9y^2}{144} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \]

### Step 2: Identify the values of \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can identify:
\[ a^2 = 9 \quad \text{and} \quad b^2 = 16 \]
Thus, \(a = 3\) and \(b = 4\).

### Step 3: Determine the slope of the given line
The line is given by \(x = 2y\), which can be rewritten as:
\[ y = \frac{1}{2}x \]
This indicates that the slope \(m_1\) of the line is \(\frac{1}{2}\).

### Step 4: Use the relationship between slopes of conjugate lines
For a hyperbola, the slopes of the asymptotes are given by the formula:
\[ m_1 \cdot m = \frac{b^2}{a^2} \]
Where \(m\) is the slope of the conjugate line. Substituting the known values:
\[ \frac{1}{2} \cdot m = \frac{16}{9} \]
Solving for \(m\):
\[ m = \frac{16}{9} \cdot 2 = \frac{32}{9} \]

### Step 5: Write the equation of the conjugate line
The equation of the line with slope \(m = \frac{32}{9}\) passing through the origin is:
\[ y = \frac{32}{9}x \]

### Step 6: Find the diameter of the hyperbola
The diameter of the hyperbola is given by the distance between the vertices along the conjugate axis, which is \(2b\):
\[ \text{Diameter} = 2b = 2 \cdot 4 = 8 \]

### Final Answer
The diameter of the hyperbola conjugate to the line \(x = 2y\) is \(8\).

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