If `x^2+alphay^2+2betay=a^2` represents a pair of perpendicular straight lines , then

Options

A. `alpha=1,beta=a`
B. `alpha=1,beta=-a`
C. `alpha=-1,beta=-a`
D. `alpha=-1,beta=a`

Answer

To solve the problem, we need to find the values of \(\alpha\) and \(\beta\) such that the equation \(x^2 + \alpha y^2 + 2\beta y = a^2\) represents a pair of perpendicular straight lines.

### Step-by-Step Solution:

1. **Start with the Given Equation:**
\[ x^2 + \alpha y^2 + 2\beta y = a^2 \]

2. **Rearrange the Equation:**
We can rearrange the equation to set it to zero:
\[ x^2 + \alpha y^2 + 2\beta y - a^2 = 0 \]

3. **Complete the Square for the \(y\) Terms:**
To complete the square for the \(y\) terms, we add and subtract \(\frac{\beta^2}{\alpha}\):
\[ x^2 + \alpha \left( y^2 + \frac{2\beta}{\alpha} y + \frac{\beta^2}{\alpha^2} - \frac{\beta^2}{\alpha^2} \right) - a^2 = 0 \]
This simplifies to:
\[ x^2 + \alpha \left( \left(y + \frac{\beta}{\alpha}\right)^2 - \frac{\beta^2}{\alpha^2} \right) - a^2 = 0 \]

4. **Rewrite the Equation:**
We can rewrite the equation as:
\[ x^2 + \alpha \left(y + \frac{\beta}{\alpha}\right)^2 - \frac{\beta^2}{\alpha} - a^2 = 0 \]
Rearranging gives:
\[ x^2 + \alpha \left(y + \frac{\beta}{\alpha}\right)^2 = a^2 + \frac{\beta^2}{\alpha} \]

5. **Identify Coefficients for the Condition of Perpendicularity:**
The general form of a conic section is \(Ax^2 + By^2 + 2Hxy + 2Gx + 2Fy + C = 0\). For our equation, we have:
- \(A = 1\)
- \(B = \alpha\)
- \(H = 0\)

The condition for perpendicularity of the lines is given by:
\[ |A + B| = 0 \]
Thus, we have:
\[ |1 + \alpha| = 0 \]

6. **Solve for \(\alpha\):**
From \(1 + \alpha = 0\), we find:
\[ \alpha = -1 \]

7. **Use the Condition for the Pair of Straight Lines:**
The second condition for the pair of straight lines is:
\[ ABC + 2HGF - AF^2 - CH^2 - BG^2 = 0 \]
Substituting \(A = 1\), \(B = -1\), \(C = -a^2\), \(H = 0\), \(G = 0\), and \(F = \beta\):
\[ 1(-1)(-a^2) + 0 - 1(\beta^2) - 0 - (-1)(0) = 0 \]
This simplifies to:
\[ a^2 - \beta^2 = 0 \]

8. **Solve for \(\beta\):**
From \(a^2 - \beta^2 = 0\), we find:
\[ \beta^2 = a^2 \implies \beta = a \]

### Final Values:
Thus, we have:
\[ \alpha = -1 \quad \text{and} \quad \beta = a \]