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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Aicharya · Accepted Answer

Let given line be $ x^2+\alpha y^2+2 \beta y-a^2=0 $ Here, $a=1, b=\alpha, h=0, g=0, f=\beta, c=-a^2$ Condition for perpendicular line $a+b=0$ $ 	herefore \quad 1+\alpha=0 \Rightarrow \alpha=-1 $ Condition of pair of lines $ \begin{array}{ccc} & a b c+2 f g h-a f^2-b g^2-c h^2=0 \ 	herefore & 1 	imes \alpha 	imes\left(-a^2ight)+0-1(\beta)^2-0-\left(-a^2ight)(0)=0 \ \Rightarrow & -a^2 \alpha-\beta^2=0 \ \Rightarrow & \beta^2=-\alpha a^2 \ \Rightarrow & \beta^2=-(-1) a^2 \quad(\because \alpha=-1) \ \Rightarrow & \beta^2=a^2 \Rightarrow \beta=a \end{array} $

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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