If `theta` is the acute angle between the lines represented by `kx^(2)-4xy+y^(2)=0 and tantheta=(1)/(2)`, then k=

Options

A. `-21, -3`
B. `-21, 3`
C. `21, -3`
D. `21, 3`

Answer

To solve the problem, we need to find the value of \( k \) given that the acute angle \( \theta \) between the lines represented by the equation \( kx^2 - 4xy + y^2 = 0 \) and \( \tan \theta = \frac{1}{2} \).

### Step-by-Step Solution:

1. **Identify the coefficients from the given equation:**
The equation of the pair of lines is given as:
\[ kx^2 - 4xy + y^2 = 0 \]
Comparing this with the standard form \( Ax^2 + By^2 + 2Hxy = 0 \), we identify:
- \( A = k \)
- \( B = 1 \)
- \( H = -2 \)

2. **Use the formula for the tangent of the angle between the lines:**
The formula for the tangent of the angle \( \theta \) between the lines is given by:
\[ \tan \theta = \frac{2\sqrt{H^2 - AB}}{A + B} \]
Substituting the values we found:
\[ \tan \theta = \frac{2\sqrt{(-2)^2 - k \cdot 1}}{k + 1} \]
Simplifying this gives:
\[ \tan \theta = \frac{2\sqrt{4 - k}}{k + 1} \]

3. **Set the expression equal to the given tangent value:**
We know from the problem that:
\[ \tan \theta = \frac{1}{2} \]
Therefore, we set up the equation:
\[ \frac{2\sqrt{4 - k}}{k + 1} = \frac{1}{2} \]

4. **Cross-multiply to eliminate the fraction:**
Cross-multiplying gives:
\[ 4\sqrt{4 - k} = k + 1 \]

5. **Square both sides to eliminate the square root:**
Squaring both sides results in:
\[ 16(4 - k) = (k + 1)^2 \]
Expanding both sides:
\[ 64 - 16k = k^2 + 2k + 1 \]

6. **Rearrange the equation:**
Bringing all terms to one side gives:
\[ k^2 + 18k - 63 = 0 \]

7. **Factor the quadratic equation:**
We can factor this quadratic as:
\[ (k + 21)(k - 3) = 0 \]

8. **Solve for \( k \):**
Setting each factor to zero gives us:
\[ k + 21 = 0 \quad \Rightarrow \quad k = -21 \]
\[ k - 3 = 0 \quad \Rightarrow \quad k = 3 \]

Thus, the values of \( k \) are \( k = 3 \) and \( k = -21 \).

### Final Answer:
The values of \( k \) are \( k = 3 \) or \( k = -21 \).