The combined equation of the asymptotes of the hyperbola `2x^2 + 5xy + 2y^2 + 4x + 5y = 0` is -

Options

A. `2x^(2)+5xy+2y^(2)+4x+5y+2=0`
B. `2x^(2)+5xy+2y^(2)+4x+5y-2=0`
C. `2x^(2)+5xy+2y^(2)=0`
D. none of these

Explanation

To find the combined equation of the asymptotes of the hyperbola given by the equation \(2x^2 + 5xy + 2y^2 + 4x + 5y = 0\), we can follow these steps:

### Step 1: Write the given equation
The given equation of the hyperbola is:
\[ 2x^2 + 5xy + 2y^2 + 4x + 5y = 0 \]

### Step 2: Formulate the combined equation of the asymptotes
The combined equation of the asymptotes can be expressed as:
\[ 2x^2 + 5xy + 2y^2 + 4x + 5y + \lambda = 0 \]
where \(\lambda\) is a constant that we need to determine.

### Step 3: Identify coefficients for the determinant condition
To find the value of \(\lambda\), we need to use the condition that the determinant (denoted as \(\Delta\)) must equal zero for the asymptotes to be straight lines. The determinant is given by:
\[ \Delta = ABC - 2FGH \]
where \(A\), \(B\), \(C\), \(F\), \(G\), and \(H\) are coefficients from the general conic section equation:
\[ Ax^2 + Bxy + Cy^2 + 2Fx + 2Gy + H = 0 \]
From our equation, we can identify:
- \(A = 2\)
- \(B = 5\)
- \(C = 2 + \lambda\)
- \(F = 2\)
- \(G = \frac{5}{2}\)
- \(H = \lambda\)

### Step 4: Substitute values into the determinant formula
Substituting these values into the determinant formula, we have:
\[ \Delta = (2)(2 + \lambda)(2) - 2(2)\left(\frac{5}{2}\right)(\lambda) \]
Expanding this gives:
\[ \Delta = 8 + 4\lambda - 10\lambda = 8 - 6\lambda \]

### Step 5: Set the determinant to zero
For the asymptotes to be straight lines, we set \(\Delta = 0\):
\[ 8 - 6\lambda = 0 \]
Solving for \(\lambda\):
\[ 6\lambda = 8 \implies \lambda = \frac{8}{6} = \frac{4}{3} \]

### Step 6: Substitute \(\lambda\) back into the combined equation
Now, substituting \(\lambda\) back into the combined equation of the asymptotes:
\[ 2x^2 + 5xy + 2y^2 + 4x + 5y + \frac{4}{3} = 0 \]
To eliminate the fraction, we can multiply the entire equation by 3:
\[ 6x^2 + 15xy + 6y^2 + 12x + 15y + 4 = 0 \]

### Final Answer
Thus, the combined equation of the asymptotes of the hyperbola is:
\[ 6x^2 + 15xy + 6y^2 + 12x + 15y + 4 = 0 \]

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