The equation to the pair of lines passing through origin and perpendicular to
`2x^(2)+5xy+2y^(2)+10x+5y=0` is …………..

Answer

To find the equation of the pair of lines passing through the origin and perpendicular to the given equation \(2x^2 + 5xy + 2y^2 + 10x + 5y = 0\), we can follow these steps:

### Step 1: Identify the coefficients
The given equation can be expressed in the general form of a conic section:
\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]
Here, we have:
- \(A = 2\)
- \(B = 5\)
- \(C = 2\)
- \(D = 10\)
- \(E = 5\)
- \(F = 0\)

### Step 2: Find the condition for perpendicularity
The condition for two lines represented by the equation \(Ax^2 + Bxy + Cy^2 = 0\) to be perpendicular is given by:
\[ B^2 - 4AC = 0 \]
For our case, we need to find the equation of lines that are perpendicular to the given lines. The slopes of the lines represented by the given equation can be found using:
\[ m^2 - \frac{B}{A}m + \frac{C}{A} = 0 \]
Thus, we will find the slopes of the lines represented by the given equation.

### Step 3: Calculate the slopes
Using the quadratic formula, the slopes \(m\) of the lines can be calculated as:
\[ m = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Substituting the values:
\[ m = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \]
\[ m = \frac{-5 \pm \sqrt{25 - 16}}{4} \]
\[ m = \frac{-5 \pm 3}{4} \]
This gives us two slopes:
\[ m_1 = \frac{-2}{4} = -\frac{1}{2}, \quad m_2 = \frac{-8}{4} = -2 \]

### Step 4: Find the slopes of the perpendicular lines
The slopes of the lines perpendicular to these will be the negative reciprocals:
\[ m_1' = 2, \quad m_2' = \frac{1}{2} \]

### Step 5: Write the equations of the lines
The equations of the lines passing through the origin with slopes \(m_1'\) and \(m_2'\) can be expressed as:
\[ y = m_1'x \quad \text{and} \quad y = m_2'x \]
Thus, we have:
\[ y = 2x \quad \text{and} \quad y = \frac{1}{2}x \]

### Step 6: Combine the equations
The combined equation of the pair of lines can be written as:
\[ y - 2x = 0 \quad \text{and} \quad 2y - x = 0 \]
Multiplying the first equation by 2 gives:
\[ 2y - 4x = 0 \]
Thus, the equation of the pair of lines can be expressed as:
\[ 2y^2 - 5xy + 2x^2 = 0 \]

### Final Answer
The equation of the pair of lines passing through the origin and perpendicular to the given equation is:
\[ 2y^2 - 5xy + 2x^2 = 0 \]