If the pairs of lines `x^(2)+2xy+ay^(2)=0andax^(2)+2xy+y^(2)=0` have exactly one line in common then the joint equation of the other two lines is given by

Options

A. `3x^(2)+8xy-3y^(2)=0`
B. `3x^(2)+10xy+3y^(2)=0`
C. `y^(2)+2xy-3x^(2)=0`
D. `x^(2)+2xy-3y^(2)=0`

Explanation

To solve the problem of finding the joint equation of the other two lines when the pairs of lines \( x^2 + 2xy + ay^2 = 0 \) and \( ax^2 + 2xy + y^2 = 0 \) have exactly one line in common, we can follow these steps:

### Step 1: Identify the common line
Let the common line be represented as \( y = mx \). This means that for both equations, substituting \( y \) with \( mx \) should yield a quadratic equation in terms of \( m \).

### Step 2: Substitute into the first equation
Substituting \( y = mx \) into the first equation \( x^2 + 2xy + ay^2 = 0 \):
\[ x^2 + 2x(mx) + a(mx)^2 = 0 \]
This simplifies to:
\[ x^2 + 2mx^2 + am^2x^2 = 0 \]
Factoring out \( x^2 \) (assuming \( x \neq 0 \)):
\[ (1 + 2m + am^2)x^2 = 0 \]
Thus, we have:
\[ am^2 + 2m + 1 = 0 \tag{1} \]

### Step 3: Substitute into the second equation
Now, substituting \( y = mx \) into the second equation \( ax^2 + 2xy + y^2 = 0 \):
\[ ax^2 + 2x(mx) + (mx)^2 = 0 \]
This simplifies to:
\[ ax^2 + 2mx^2 + m^2x^2 = 0 \]
Factoring out \( x^2 \):
\[ (ax + 2m + m^2)x^2 = 0 \]
Thus, we have:
\[ m^2 + 2m + a = 0 \tag{2} \]

### Step 4: Set the equations equal
Since both equations (1) and (2) must have the same root \( m \), we can set them equal to each other. From equations (1) and (2):
\[ am^2 + 2m + 1 = 0 \]
\[ m^2 + 2m + a = 0 \]

### Step 5: Eliminate \( m \)
To find the condition for \( a \), we can eliminate \( m \) from these equations. We can express \( m^2 \) from both equations and set them equal:
From (1):
\[ m^2 = \frac{-2m - 1}{a} \]
From (2):
\[ m^2 = -2m - a \]
Setting these equal gives:
\[ \frac{-2m - 1}{a} = -2m - a \]
Cross-multiplying leads to:
\[ -2m - 1 = -2am + a^2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ a^2 - 2am + 2m + 1 = 0 \]

### Step 7: Discriminant condition
For the lines to have exactly one line in common, the discriminant of this quadratic in \( m \) must be zero:
\[ D = (2a - 2)^2 - 4a^2 = 0 \]
Solving this gives:
\[ 4a - 4 = 0 \implies a = -3 \]

### Step 8: Finding the joint equation of the other two lines
Substituting \( a = -3 \) back into the original equations gives:
\[ x^2 + 2xy - 3y^2 = 0 \quad \text{and} \quad -3x^2 + 2xy + y^2 = 0 \]
Factoring these equations leads to:
1. From \( x^2 + 2xy - 3y^2 = 0 \):
\[ (x - y)(x + 3y) = 0 \]
2. From \( -3x^2 + 2xy + y^2 = 0 \):
\[ -(x - y)(3x + y) = 0 \]

### Final Joint Equation
The joint equation of the other two lines is:
\[ (x - y)(x + 3y)(3x + y) = 0 \]