Two pairs of straight lines have the equations `y^(2)+xy-12x^(2)=0andax^(2)+2hxy+by^(2)=0`. One line will be common among them if

Options

A. `a+8h-16b=0`
B. `a-8h+16b=0`
C. `a-6h+9b=0`
D. `a+6h+9b=0`

Explanation

To determine the condition under which one line will be common among the two pairs of straight lines given by the equations \( y^2 + xy - 12x^2 = 0 \) and \( ax^2 + 2hxy + by^2 = 0 \), we can follow these steps:

### Step 1: Factor the first equation
The first equation is:
\[ y^2 + xy - 12x^2 = 0 \]
We can factor this equation. We rewrite it as:
\[ y^2 + 4xy - 3xy - 12x^2 = 0 \]
Now, we can group the terms:
\[ y(y + 4x) - 3x(y + 4x) = 0 \]
This gives us:
\[ (y + 4x)(y - 3x) = 0 \]
Thus, the two lines represented by the first equation are:
\[ y + 4x = 0 \quad \text{and} \quad y - 3x = 0 \]

### Step 2: Find slopes of the lines
From the factored form, we can find the slopes of the lines:
1. From \( y + 4x = 0 \), we have \( y = -4x \) (slope = -4).
2. From \( y - 3x = 0 \), we have \( y = 3x \) (slope = 3).

### Step 3: Analyze the second equation
The second equation is:
\[ ax^2 + 2hxy + by^2 = 0 \]
This is a general form of a pair of straight lines. The condition for this equation to represent two lines is that the determinant of the coefficients must be zero:
\[ D = \begin{vmatrix} a & h \\ h & b \end{vmatrix} = ab - h^2 = 0 \]

### Step 4: Condition for common line
For one line to be common between the two pairs, the slopes found from the first equation must satisfy the second equation. Therefore, we substitute the slopes into the second equation.

1. For \( y = 3x \) (slope = 3):
\[ b(3^2) + 2h(3) + a = 0 \implies 9b + 6h + a = 0 \]

2. For \( y = -4x \) (slope = -4):
\[ b(-4^2) + 2h(-4) + a = 0 \implies 16b - 8h + a = 0 \]

### Step 5: Set up the equations
Now we have two equations:
1. \( 9b + 6h + a = 0 \)
2. \( 16b - 8h + a = 0 \)

### Step 6: Solve the system of equations
To find the condition for \( a \), \( b \), and \( h \), we can eliminate \( a \) from both equations:
From the first equation, we have:
\[ a = -9b - 6h \]
Substituting this into the second equation:
\[ 16b - 8h - 9b - 6h = 0 \]
This simplifies to:
\[ 7b - 14h = 0 \implies b = 2h \]

### Step 7: Substitute back to find \( a \)
Now substituting \( b = 2h \) back into the equation for \( a \):
\[ a = -9(2h) - 6h = -18h - 6h = -24h \]

### Conclusion
Thus, the condition for one line to be common among the two pairs of straight lines is:
\[ a = -24h \quad \text{and} \quad b = 2h \]