Question

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

• A. 4 a
• B. a
• C. 2 a
• D. 3 a

Answer 1

The Correct Option is B

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 $a+b=0$ $\therefore 1+\alpha=0$ $ \Rightarrow \alpha=-1$ Condition of pair of lines $a b c+2 f g h-a f^{2}-b g^{2}-c h^{2}=0$ $\therefore 1 \times \alpha \times\left(-a^{2}\right)+0-1(\beta)^{2}-0-\left(-a^{2}\right)(0)=0$ $\Rightarrow -a^{2} \alpha-\beta^{2}=0$ $\Rightarrow \beta^{2}=-\alpha a^{2}$ $\Rightarrow \beta^{2}=-(-1) a^{2} \,\,\,(\because \alpha=-1)$ $\Rightarrow \beta^{2}=a^{2} $ $\Rightarrow \beta=a$