Find the separate equation of two straight lines whose joint equation is ab `(x^2-y^2) +(a^2-b^2)xy=0`

Answer

To find the separate equations of two straight lines whose joint equation is given by:

\[ ab (x^2 - y^2) + (a^2 - b^2)xy = 0 \]

we can follow these steps:

### Step 1: Rewrite the Joint Equation
Start with the given joint equation:

\[ ab (x^2 - y^2) + (a^2 - b^2)xy = 0 \]

### Step 2: Expand the Equation
We can rewrite the equation as:

\[ ab x^2 - ab y^2 + (a^2 - b^2)xy = 0 \]

### Step 3: Divide by \( x^2 \)
To simplify, divide the entire equation by \( x^2 \):

\[ ab - ab \left(\frac{y}{x}\right)^2 + (a^2 - b^2) \left(\frac{y}{x}\right) = 0 \]

Let \( m = \frac{y}{x} \), then the equation becomes:

\[ ab - ab m^2 + (a^2 - b^2)m = 0 \]

### Step 4: Rearrange the Equation
Rearranging gives us a quadratic equation in terms of \( m \):

\[ -ab m^2 + (a^2 - b^2)m + ab = 0 \]

### Step 5: Apply the Quadratic Formula
Using the quadratic formula \( m = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), where \( A = -ab \), \( B = a^2 - b^2 \), and \( C = ab \):

1. Calculate \( B^2 - 4AC \):
\[ (a^2 - b^2)^2 - 4(-ab)(ab) = (a^2 - b^2)^2 + 4a^2b^2 \]

2. This simplifies to:
\[ a^4 - 2a^2b^2 + b^4 + 4a^2b^2 = a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2 \]

3. Now substitute back into the quadratic formula:
\[ m = \frac{-(a^2 - b^2) \pm (a^2 + b^2)}{-2ab} \]

### Step 6: Solve for the Slopes
This gives us two solutions for \( m \):
1. \( m_1 = \frac{2b^2}{2ab} = \frac{b}{a} \)
2. \( m_2 = \frac{0}{-2ab} = 0 \)

### Step 7: Write the Equations of the Lines
The slopes correspond to the equations of the lines:
1. For \( m_1 = \frac{b}{a} \):
\[ y = \frac{b}{a}x \implies ax - by = 0 \]

2. For \( m_2 = 0 \):
\[ y = 0 \implies b x + 0y = 0 \]

Thus, the separate equations of the two straight lines are:

1. \( ax - by = 0 \)
2. \( bx + ay = 0 \)

### Final Answer
The separate equations of the two straight lines are:
1. \( ax - by = 0 \)
2. \( bx + ay = 0 \)

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