The eqautions of lines `L_(1) and L_(2)` are `y=mx and y=nx`, respectively. Suppose `L_(1)` makes twice as large an angle with the horizontal (measured counterclockwise from the positive x - axis) as does `L_(2)` and m = 4n, then the value of `((m^(2)+4n^(2)))/((m^(2)-6n^(2)))` is equal to (where, `n ne 0`)

Options

A. 3
B. `-3`
C. 2
D. `-2`

Answer

To solve the problem, we will follow these steps:

1. **Identify the angles made by the lines**:
The angle \( \theta_1 \) made by line \( L_1 \) with the horizontal is given by \( \tan(\theta_1) = m \). Similarly, for line \( L_2 \), \( \tan(\theta_2) = n \). According to the problem, \( \theta_1 = 2\theta_2 \).

2. **Use the tangent double angle formula**:
Since \( \theta_1 = 2\theta_2 \), we can use the double angle formula for tangent:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
Applying this to our angles:
\[ m = \tan(2\theta_2) = \frac{2n}{1 - n^2} \]

3. **Given relationship between m and n**:
We are also given that \( m = 4n \). Therefore, we can set up the equation:
\[ 4n = \frac{2n}{1 - n^2} \]

4. **Cross-multiply to solve for n**:
\[ 4n(1 - n^2) = 2n \]
Simplifying this gives:
\[ 4n - 4n^3 = 2n \]
Rearranging leads to:
\[ 4n^3 - 2n = 0 \]
Factoring out \( 2n \):
\[ 2n(2n^2 - 1) = 0 \]
This gives us \( n = 0 \) or \( n^2 = \frac{1}{2} \). Since \( n \neq 0 \), we have \( n = \frac{1}{\sqrt{2}} \).

5. **Find m using the relationship \( m = 4n \)**:
Substituting \( n \) back into the equation for \( m \):
\[ m = 4 \cdot \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

6. **Calculate the value of the expression**:
We need to find:
\[ \frac{m^2 + 4n^2}{m^2 - 6n^2} \]
First, calculate \( m^2 \) and \( n^2 \):
\[ m^2 = (2\sqrt{2})^2 = 8, \quad n^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
Now substitute these values into the expression:
\[ \frac{8 + 4 \cdot \frac{1}{2}}{8 - 6 \cdot \frac{1}{2}} = \frac{8 + 2}{8 - 3} = \frac{10}{5} = 2 \]

Thus, the final answer is:
\[ \boxed{2} \]