If the equation `x^(4)+bx^(3)y+cx^(2)y^(2)+dxy^(3)+ey^(4)=0` represent two pairs of perpendicular lines, then

Options

A. `b+d=1 and e=-1`
B. `b+d=0 and e=-1`
C. `b+d=0 and e=1`
D. none of these

Explanation

To solve the problem, we need to determine the conditions under which the equation \( x^4 + bx^3y + cx^2y^2 + dxy^3 + ey^4 = 0 \) represents two pairs of perpendicular lines.

### Step-by-Step Solution:

1. **Understanding the Equation**:
The given equation is a quartic equation in two variables \( x \) and \( y \). For it to represent two pairs of perpendicular lines, it can be factored into the product of two quadratic equations.

2. **Assuming the Form of the Quadratics**:
We can assume that the equation can be expressed as:
\[ (x^2 + \alpha xy - y^2)(x^2 + \beta xy - y^2) = 0 \]
where \( \alpha \) and \( \beta \) are constants.

3. **Expanding the Product**:
Expanding the product gives:
\[ x^4 + (\alpha + \beta)x^3y + (\alpha\beta - 2)x^2y^2 + (-\alpha - \beta)xy^3 + y^4 = 0 \]

4. **Comparing Coefficients**:
Now we compare the coefficients of the expanded form with the original equation:
- Coefficient of \( x^3y \): \( b = \alpha + \beta \)
- Coefficient of \( x^2y^2 \): \( c = \alpha\beta - 2 \)
- Coefficient of \( xy^3 \): \( d = -(\alpha + \beta) \)
- Coefficient of \( y^4 \): \( e = 1 \)

5. **Conditions for Perpendicular Lines**:
For the lines to be perpendicular, the product of the slopes of the two lines must equal -1. This leads to the condition:
\[ \alpha \beta = -1 \]

6. **Using the Conditions**:
From the equations derived:
- Since \( e = 1 \), we have \( e = 1 \).
- From \( d = -(\alpha + \beta) \) and \( b = \alpha + \beta \), we can conclude:
\[ b + d = 0 \]

7. **Final Conditions**:
Thus, the conditions we have derived are:
- \( e = 1 \)
- \( b + d = 0 \)
- \( \alpha \beta = -1 \)

### Summary of Results:
The equation \( x^4 + bx^3y + cx^2y^2 + dxy^3 + ey^4 = 0 \) represents two pairs of perpendicular lines if:
- \( e = 1 \)
- \( b + d = 0 \)