If acute angle between lines `3x^(2)-4xy+by^(2)=0` is `cot^(-1)2,` then `b=`

Options

A. `1,-55`
B. `-1,55`
C. `15,-5`
D. `1,-54`

Answer

To solve the problem, we need to find the value of \( b \) such that the acute angle between the lines represented by the equation \( 3x^2 - 4xy + by^2 = 0 \) is \( \cot^{-1}(2) \).

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given equation is in the form \( ax^2 + 2hxy + by^2 = 0 \).
Here, \( a = 3 \), \( 2h = -4 \) (thus \( h = -2 \)), and \( b = b \).

2. **Use the formula for the angle between two lines**:
The formula for the tangent of the angle \( \theta \) between two lines is given by:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]
We know that \( \theta = \cot^{-1}(2) \), which means:
\[ \cot \theta = 2 \quad \Rightarrow \quad \tan \theta = \frac{1}{2} \]

3. **Set up the equation**:
Using the formula:
\[ \frac{1}{2} = \frac{2\sqrt{(-2)^2 - (3)(b)}}{3 + b} \]

4. **Cross-multiply to eliminate the fraction**:
\[ 1(3 + b) = 4\sqrt{4 - 3b} \]
This simplifies to:
\[ 3 + b = 4\sqrt{4 - 3b} \]

5. **Square both sides to eliminate the square root**:
\[ (3 + b)^2 = 16(4 - 3b) \]
Expanding both sides gives:
\[ 9 + 6b + b^2 = 64 - 48b \]

6. **Rearrange the equation**:
Bringing all terms to one side results in:
\[ b^2 + 54b - 55 = 0 \]

7. **Factor the quadratic equation**:
We can factor this as:
\[ (b + 55)(b - 1) = 0 \]

8. **Find the values of \( b \)**:
Setting each factor to zero gives:
\[ b + 55 = 0 \quad \Rightarrow \quad b = -55 \]
\[ b - 1 = 0 \quad \Rightarrow \quad b = 1 \]

### Final Answer:
Thus, the values of \( b \) are \( b = 1 \) and \( b = -55 \).

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