The equation `8x^(2)+8xy+2y^(2)+26x+13y+15=0` represents a pair of parallel straight lines. The distance between them is

Options

A. `7/(sqrt(5))`
B. `(sqrt(7))/5`
C. `7/(2sqrt(5))`
D. none of these

Answer

To find the distance between the parallel lines represented by the equation \(8x^2 + 8xy + 2y^2 + 26x + 13y + 15 = 0\), we will follow these steps:

### Step 1: Identify the coefficients
The given equation can be compared to the general form of a conic section:
\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]
From the equation, we can identify:
- \(a = 8\)
- \(b = 2\)
- \(h = 4\)
- \(g = 13\)
- \(f = \frac{13}{2}\)
- \(c = 15\)

### Step 2: Check the condition for parallel lines
For the equation to represent a pair of parallel lines, the condition \(h^2 - ab = 0\) must hold.
Calculating:
\[ h^2 - ab = 4^2 - (8)(2) = 16 - 16 = 0 \]
Since this condition is satisfied, the equation represents a pair of parallel lines.

### Step 3: Simplify the equation
To make calculations easier, divide the entire equation by 2:
\[ 4x^2 + 4xy + y^2 + 13x + \frac{13}{2}y + \frac{15}{2} = 0 \]

### Step 4: Factor the equation
Assuming the factors of the equation are of the form:
\[ (2x + y + p)(2x + y + q) = 0 \]
Expanding this gives:
\[ 4x^2 + 2(2p + 2q)x + (pq)y + (p + q)y + pq = 0 \]
From this, we can derive the following equations:
1. \(2(p + q) = 13\)
2. \(p + q = \frac{13}{2}\)
3. \(pq = \frac{15}{2}\)

### Step 5: Solve for \(p\) and \(q\)
From \(p + q = \frac{13}{2}\), we can express \(q\) in terms of \(p\):
\[ q = \frac{13}{2} - p \]
Substituting into \(pq = \frac{15}{2}\):
\[ p\left(\frac{13}{2} - p\right) = \frac{15}{2} \]
Expanding and rearranging gives:
\[ -p^2 + \frac{13}{2}p - \frac{15}{2} = 0 \]
Multiplying through by -2 to eliminate fractions:
\[ 2p^2 - 13p + 15 = 0 \]

### Step 6: Factor the quadratic equation
Factoring gives:
\[ (2p - 5)(p - 3) = 0 \]
Thus, \(p = \frac{5}{2}\) or \(p = 3\).

### Step 7: Find corresponding \(q\) values
If \(p = 5\), then \(q = \frac{13}{2} - 5 = \frac{3}{2}\).
If \(p = \frac{3}{2}\), then \(q = 5\).

### Step 8: Write the equations of the lines
The equations of the lines are:
1. \(2x + y + 5 = 0\)
2. \(2x + y + \frac{3}{2} = 0\)

### Step 9: Calculate the distance between the parallel lines
The distance \(d\) between two parallel lines of the form \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) is given by:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
Here, \(C_1 = 5\) and \(C_2 = \frac{3}{2}\):
\[ d = \frac{\left|5 - \frac{3}{2}\right|}{\sqrt{2^2 + 1^2}} = \frac{\left| \frac{10}{2} - \frac{3}{2} \right|}{\sqrt{4 + 1}} = \frac{\left| \frac{7}{2} \right|}{\sqrt{5}} = \frac{7}{2\sqrt{5}} \]

### Final Answer
The distance between the parallel lines is:
\[ \frac{7}{2\sqrt{5}} \]