If the lines represented by `x^(2)-4xy+y^(2)=0` makes angle `alpha and beta` with X-axis, then `tan^(2)alpha+tan^(2)beta=`

Options

A. `2`
B. `-2`
C. `14`
D. `-14`

Answer

To solve the problem, we need to find the value of \( \tan^2 \alpha + \tan^2 \beta \) where the lines represented by the equation \( x^2 - 4xy + y^2 = 0 \) make angles \( \alpha \) and \( \beta \) with the x-axis.

### Step-by-Step Solution:

1. **Identify the Equation**:
We start with the given equation of the lines:
\[ x^2 - 4xy + y^2 = 0 \]

2. **Rearranging the Equation**:
We can rewrite the equation in terms of \( y \) by dividing through by \( x^2 \):
\[ \frac{y^2}{x^2} - 4\frac{y}{x} + 1 = 0 \]
Letting \( m = \frac{y}{x} \), we can rewrite the equation as:
\[ m^2 - 4m + 1 = 0 \]

3. **Finding the Slopes**:
The roots of the quadratic equation \( m^2 - 4m + 1 = 0 \) will give us the slopes \( m_1 \) and \( m_2 \) of the lines:
Using the quadratic formula:
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -4, c = 1 \):
\[ m = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]
Thus, \( m_1 = 2 + \sqrt{3} \) and \( m_2 = 2 - \sqrt{3} \).

4. **Calculating \( \tan^2 \alpha + \tan^2 \beta \)**:
We need to find \( \tan^2 \alpha + \tan^2 \beta \), which is equal to \( m_1^2 + m_2^2 \):
\[ m_1^2 + m_2^2 = (m_1 + m_2)^2 - 2m_1 m_2 \]

5. **Finding \( m_1 + m_2 \) and \( m_1 m_2 \)**:
From the properties of quadratic equations, we know:
\[ m_1 + m_2 = 4 \quad \text{(sum of roots)} \]
\[ m_1 m_2 = 1 \quad \text{(product of roots)} \]

6. **Substituting Values**:
Now substituting these values into the equation:
\[ m_1^2 + m_2^2 = (m_1 + m_2)^2 - 2m_1 m_2 = 4^2 - 2 \cdot 1 = 16 - 2 = 14 \]

7. **Final Answer**:
Therefore, the value of \( \tan^2 \alpha + \tan^2 \beta \) is:
\[ \boxed{14} \]