The value of k for which the equation `2x^2+5y^2-2ky+4x+6y=0` represent the pair of straight lines is

Options

A. `-19/6`
B. `19/6`
C. `-7/6`
D. none of these

Answer

To find the value of \( k \) for which the equation \( 2x^2 + 5y^2 - 2ky + 4x + 6y = 0 \) represents a pair of straight lines, we can follow these steps:

### Step 1: Identify coefficients
The general form of the equation 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:
- \( a = 2 \)
- \( b = 5 \)
- \( 2h = 0 \) (since there is no \( xy \) term, \( h = 0 \))
- \( 2g = 4 \) → \( g = 2 \)
- \( 2f = 6 \) → \( f = 3 \)
- \( c = 0 \)

### Step 2: Write the determinant condition
For the equation to represent a pair of straight lines, the determinant \( \Delta \) must be zero. The determinant is given by:
\[ \Delta = ABC + 2FGH - AF^2 - BG^2 - CH^2 \]
Substituting the values we have:
- \( A = a = 2 \)
- \( B = b = 5 \)
- \( C = c = 0 \)
- \( F = f = 3 \)
- \( G = g = 2 \)
- \( H = h = 0 \)

### Step 3: Substitute values into the determinant
Now substituting these values into the determinant condition:
\[ \Delta = 2 \cdot 5 \cdot 0 + 2 \cdot 3 \cdot 2 \cdot 0 - 2 \cdot 3^2 - 5 \cdot 2^2 - 0 \]
This simplifies to:
\[ \Delta = 0 + 0 - 2 \cdot 9 - 5 \cdot 4 \]
\[ \Delta = -18 - 20 = -38 \]

### Step 4: Include \( k \) in the determinant
We have \( H = -k \), so we need to include that in our determinant:
\[ \Delta = 2 \cdot 5 \cdot 0 + 2 \cdot 3 \cdot 2 \cdot (-k) - 2 \cdot 3^2 - 5 \cdot 2^2 - 0 \]
This gives:
\[ \Delta = 0 - 12k - 18 - 20 \]
Setting \( \Delta = 0 \):
\[ -12k - 38 = 0 \]

### Step 5: Solve for \( k \)
Now, we can solve for \( k \):
\[ -12k = 38 \]
\[ k = -\frac{38}{12} = -\frac{19}{6} \]

### Final Answer
Thus, the value of \( k \) for which the equation represents a pair of straight lines is:
\[ \boxed{-\frac{19}{6}} \]