`9x^(2) +2hxy +4y^(2) +6x +2fy - 3 = 0` represents two parallel lines. Then

Options

A. `h = 6, f = 2`
B. `h =- 6, f = 2`
C. `h = 6, f =- 2`
D. `h =- 6, f =- 2`

Explanation

To solve the problem, we need to determine the conditions under which the given equation represents two parallel lines. The given equation is:

\[ 9x^2 + 2hxy + 4y^2 + 6x + 2fy - 3 = 0 \]

### Step 1: Identify coefficients
The general form of a pair of straight lines is given by:

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]

From the given equation, we can identify the coefficients:
- \( a = 9 \)
- \( b = 4 \)
- \( h = h \)
- \( g = 3 \)
- \( f = f \)
- \( c = -3 \)

### Step 2: Condition for parallel lines
For the lines to be parallel, the condition is:

\[ h^2 = ab \]

Substituting the values of \( a \) and \( b \):

\[ h^2 = 9 \times 4 = 36 \]

### Step 3: Solve for \( h \)
Taking the square root of both sides gives:

\[ h = \pm 6 \]

### Step 4: Condition for pair of lines
Next, we need to ensure that the given equation represents a pair of straight lines. This is determined by the determinant condition:

\[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \]

Substituting the values:

\[ \begin{vmatrix} 9 & h & 3 \\ h & 4 & f \\ 3 & f & -3 \end{vmatrix} = 0 \]

### Step 5: Calculate the determinant
Expanding the determinant along the first column:

\[ = 9 \begin{vmatrix} 4 & f \\ f & -3 \end{vmatrix} - h \begin{vmatrix} h & f \\ 3 & -3 \end{vmatrix} + 3 \begin{vmatrix} h & 4 \\ 3 & f \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 4 & f \\ f & -3 \end{vmatrix} = 4(-3) - f^2 = -12 - f^2 \)

2. \( \begin{vmatrix} h & f \\ 3 & -3 \end{vmatrix} = h(-3) - 3f = -3h - 3f \)

3. \( \begin{vmatrix} h & 4 \\ 3 & f \end{vmatrix} = hf - 12 \)

Substituting back into the determinant:

\[ 9(-12 - f^2) - h(-3h - 3f) + 3(hf - 12) = 0 \]

Expanding this gives:

\[ -108 - 9f^2 + 3h^2 + 3hf + 3hf - 36 = 0 \]

Combining like terms:

\[ -9f^2 + 6hf + 3h^2 - 144 = 0 \]

### Step 6: Solve for \( f \)
Now we can solve this quadratic equation for \( f \):

\[ 9f^2 - 6hf + (144 - 3h^2) = 0 \]

Using the quadratic formula \( f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 9 \), \( b = -6h \), and \( c = 144 - 3h^2 \).

### Step 7: Substitute \( h = 6 \) and \( h = -6 \)
1. For \( h = 6 \):

\[ 9f^2 - 36f + (144 - 108) = 0 \implies 9f^2 - 36f + 36 = 0 \]
Dividing through by 9:
\[ f^2 - 4f + 4 = 0 \implies (f - 2)^2 = 0 \implies f = 2 \]

2. For \( h = -6 \):

\[ 9f^2 + 36f + (144 - 108) = 0 \implies 9f^2 + 36f + 36 = 0 \]
Dividing through by 9:
\[ f^2 + 4f + 4 = 0 \implies (f + 2)^2 = 0 \implies f = -2 \]

### Final Result
Thus, the values of \( f \) corresponding to \( h = 6 \) and \( h = -6 \) are:

- If \( h = 6 \), then \( f = 2 \)
- If \( h = -6 \), then \( f = -2 \)