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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 second-degree curve $Ax^2 + 2Hxy + By

Vedclass · Accepted Answer

$a$

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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 second-degree curve $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$,we get $A=1, B=\alpha, H=0, G=0, F=\beta, C=-a^2$.For the lines to be perpendicular,the condition is $A + B = 0$.Therefore,$1 + \alpha = 0 \Rightarrow \alpha = -1$.The condition for the equation to represent a pair of lines is $ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0$.Substituting the values: $(1)(\alpha)(-a^2) + 2(\beta)(0)(0) - (1)(\beta)^2 - (\alpha)(0)^2 - (-a^2)(0)^2 = 0$.This simplifies to $-\alpha a^2 - \beta^2 = 0$.Substituting $\alpha = -1$,we get $-(-1)a^2 - \beta^2 = 0$,which implies $a^2 - \beta^2 = 0$.Thus,$\beta^2 = a^2$,which gives $\beta = \pm a$. Given the options,$\beta = a$ is the correct choice.

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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