If x^2+alpha y^2+2 beta y=a^2 represents a pa | vedclass

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(B) The given equation is $x^2 + \alpha y^2 + 2 \beta y - a^2 = 0$. Comparing this with the general equation of a pair of lines $Ax^2 + 2Hxy + By^2 +

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$a$

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(B) The given equation is $x^2 + \alpha y^2 + 2 \beta y - a^2 = 0$. Comparing this with the general equation of a pair of lines $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$,we get: $A = 1, B = \alpha, H = 0, G = 0, F = \beta, C = -a^2$. For the lines to be perpendicular,the condition is $A + B = 0$. $1 + \alpha = 0 \Rightarrow \alpha = -1$. For the equation to represent a pair of lines,the determinant condition must be satisfied: $ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0$. Substituting the values: $(1)(\alpha)(-a^2) + 2(\beta)(0)(0) - (1)(\beta)^2 - (\alpha)(0)^2 - (-a^2)(0)^2 = 0$. $-\alpha a^2 - \beta^2 = 0$. Since $\alpha = -1$,we have: $-(-1)a^2 - \beta^2 = 0 $ $\Rightarrow a^2 - \beta^2 = 0 $ $\Rightarrow \beta^2 = a^2 $ $\Rightarrow \beta = a$ (taking the positive root as per options).

If $x^2+\alpha y^2+2 \beta y=a^2$ represents a pair of perpendicular lines,then $\beta$ equals to

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