If `theta` is the acute angle between the lines given by `x^(2)-2pxy+y^(2)=0`, then

Options

A. `cos theta=p`
B. `tan theta=p`
C. `sec theta = p`
D. `cot theta=p`

Explanation

To solve the problem, we need to find the relationship between the acute angle \( \theta \) between the lines given by the equation \( x^2 - 2pxy + y^2 = 0 \) and the parameter \( p \).

### Step-by-Step Solution:

1. **Identify the coefficients**: The given equation is \( x^2 - 2pxy + y^2 = 0 \). We can rewrite it in the standard form of a pair of straight lines \( Ax^2 + Bxy + Cy^2 = 0 \), where:
- \( A = 1 \)
- \( B = -2p \)
- \( C = 1 \)

2. **Use the formula for the tangent of the angle between two lines**: The formula for the tangent of the angle \( \theta \) between two lines represented by the equation \( Ax^2 + Bxy + Cy^2 = 0 \) is given by:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + c} \]
where \( h = \frac{B}{2} \), \( a = A \), and \( c = C \).

3. **Calculate \( h \)**: From our coefficients:
\[ h = \frac{-2p}{2} = -p \]

4. **Calculate \( ab \)**: We have:
\[ ab = A \cdot C = 1 \cdot 1 = 1 \]

5. **Substitute into the formula**:
\[ \tan \theta = \frac{2\sqrt{(-p)^2 - 1}}{1 + 1} = \frac{2\sqrt{p^2 - 1}}{2} = \sqrt{p^2 - 1} \]

6. **Relate \( \tan \theta \) to \( p \)**: We know that:
\[ \tan^2 \theta + 1 = \sec^2 \theta \]
Therefore:
\[ 1 + \tan^2 \theta = 1 + (p^2 - 1) = p^2 \]
This implies:
\[ \sec^2 \theta = p^2 \]

7. **Find \( \sec \theta \)**: Since \( \sec^2 \theta = p^2 \), we take the positive root (since \( \theta \) is acute):
\[ \sec \theta = p \]

### Conclusion:
The relationship between the acute angle \( \theta \) and the parameter \( p \) is:
\[ \sec \theta = p \]