If acute angle between lines `x^(2)-2hxy+y^(2)=0` is `60^(@)` then `h=`

Options

A. `-2`
B. `+-2`
C. `2`
D. `sqrt(3)`

Answer

To solve the problem, we need to find the value of \( h \) given that the acute angle between the lines represented by the equation \( x^2 - 2hx + y^2 = 0 \) is \( 60^\circ \).

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given equation is \( x^2 - 2hxy + y^2 = 0 \). We can compare this with the general form \( ax^2 + 2hxy + by^2 = 0 \). Here, we have:
- \( a = 1 \)
- \( b = 1 \)
- \( 2h = -2h \) (we will use \( h \) directly in the formula)

2. **Use the formula for the angle between two lines**:
The formula for the tangent of the angle \( \theta \) between two lines represented by the equation \( ax^2 + 2hxy + by^2 = 0 \) is given by:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]
In our case, substituting \( a = 1 \), \( b = 1 \):
\[ \tan \theta = \frac{2\sqrt{h^2 - 1}}{1 + 1} = \frac{2\sqrt{h^2 - 1}}{2} = \sqrt{h^2 - 1} \]

3. **Set up the equation using the given angle**:
We know that the acute angle \( \theta \) is \( 60^\circ \). Therefore:
\[ \tan 60^\circ = \sqrt{3} \]
Thus, we can set up the equation:
\[ \sqrt{h^2 - 1} = \sqrt{3} \]

4. **Square both sides**:
To eliminate the square root, we square both sides:
\[ h^2 - 1 = 3 \]

5. **Solve for \( h^2 \)**:
Rearranging the equation gives:
\[ h^2 = 3 + 1 = 4 \]

6. **Find \( h \)**:
Taking the square root of both sides, we find:
\[ h = \pm 2 \]

### Final Answer:
Thus, the values of \( h \) are \( h = 2 \) or \( h = -2 \).

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