Question

If $x^2 \alpha+y^2 +2\beta y =a^2$ represents a pair of perpendicular lines, then $|\frac{\beta}{\alpha}|=$

• A. 0
• B. 1
• C. a
• D. $\frac{a^2}{\beta}$

Answer 1

Answer: (A)

0

Answer 2

The general equation of a second-degree curve is $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. The given equation is $\alpha x^2 + y^2 + 2\beta y - a^2 = 0$. Comparing the coefficients, we have: $A = \alpha$ $H = 0$ $B = 1$ $G = 0$ $F = \beta$ $C = -a^2$

For the equation to represent a pair of perpendicular lines, the condition is $A + B = 0$. Substituting the values of $A$ and $B$: $\alpha + 1 = 0$ $\implies \alpha = -1$.

For the equation to represent a pair of lines, the discriminant must be zero: $\Delta = ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0$. Substituting the coefficients: $\Delta = (\alpha)(1)(-a^2) + 2(\beta)(0)(0) - (\alpha)(\beta)^2 - (1)(0)^2 - (-a^2)(0)^2 = 0$ $\Delta = -\alpha a^2 - \alpha \beta^2 = 0$ $\Delta = -\alpha (a^2 + \beta^2) = 0$.

Since $\alpha = -1$, we substitute this into the discriminant equation: $-(-1) (a^2 + \beta^2) = 0$ $1 (a^2 + \beta^2) = 0$ $a^2 + \beta^2 = 0$.

Assuming $a$ and $\beta$ are real numbers, the equation $a^2 + \beta^2 = 0$ implies that $a^2 = 0$ and $\beta^2 = 0$. This means $a = 0$ and $\beta = 0$.

The question asks for the value of $|\frac{\beta}{\alpha}|$. $|\frac{\beta}{\alpha}| = |\frac{0}{-1}| = 0$.