If the slope of the line given by `a^2x^2+2hxy+b^2y^2=0` be three times of the other , then h is equal to

Options

A. (a) `2sqrt(3)ab`
B. (b) `-2sqrt(3)ab`
C. (c) `(2)/(sqrt(3))ab`
D. (d) `-(2)/(sqrt(3)) ab`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Identify the equation of the pair of lines
The given equation is:
\[ a^2x^2 + 2hxy + b^2y^2 = 0 \]

This represents a pair of straight lines.

### Step 2: Use the properties of slopes
For a pair of lines given by the equation \( Ax^2 + 2Bxy + Cy^2 = 0 \), the slopes \( m_1 \) and \( m_2 \) can be found using the formulas:
1. \( m_1 + m_2 = -\frac{2h}{b^2} \)
2. \( m_1 \cdot m_2 = \frac{a^2}{b^2} \)

### Step 3: Set up the relationship between the slopes
According to the problem, one slope is three times the other:
\[ m_2 = 3m_1 \]

### Step 4: Substitute \( m_2 \) in the sum of slopes equation
Substituting \( m_2 \) into the first equation:
\[ m_1 + 3m_1 = -\frac{2h}{b^2} \]
This simplifies to:
\[ 4m_1 = -\frac{2h}{b^2} \]
From this, we can express \( m_1 \):
\[ m_1 = -\frac{h}{2b^2} \]

### Step 5: Substitute \( m_1 \) into the product of slopes equation
Now, substituting \( m_1 \) into the second equation:
\[ m_1 \cdot m_2 = \frac{a^2}{b^2} \]
Substituting \( m_2 = 3m_1 \):
\[ m_1 \cdot (3m_1) = \frac{a^2}{b^2} \]
This gives:
\[ 3m_1^2 = \frac{a^2}{b^2} \]

### Step 6: Substitute the value of \( m_1 \)
Now substitute \( m_1 = -\frac{h}{2b^2} \):
\[ 3\left(-\frac{h}{2b^2}\right)^2 = \frac{a^2}{b^2} \]
This simplifies to:
\[ 3\left(\frac{h^2}{4b^4}\right) = \frac{a^2}{b^2} \]
Multiplying both sides by \( 4b^4 \):
\[ 3h^2 = 4a^2b^2 \]

### Step 7: Solve for \( h \)
Now, we can solve for \( h \):
\[ h^2 = \frac{4a^2b^2}{3} \]
Taking the square root:
\[ h = \frac{2ab}{\sqrt{3}} \]

### Final Answer
Thus, the value of \( h \) is:
\[ h = \frac{2ab}{\sqrt{3}} \]

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