Question

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

• A. $4a$
• B. $a$
• C. $2a$
• D. $3a$

Answer 1

The correct option is B $a$

$x^2+\alpha y^2+2\beta y=a^2$
Adding and subtracting $\frac{{\beta }^2}{\alpha }$from the above equation,

$\Rightarrow x^2+\alpha y^2+2\beta y+\frac{{\beta }^2}{\alpha }-\frac{{\beta }^2}{\alpha }-a^2=0$

$\Rightarrow x^2+(\sqrt{\alpha }y+\frac{\beta }{\sqrt{\alpha }}{)}^2=a^2+\frac{{\beta }^2}{\alpha }$

For pair of straight lines $a{x}^2+b{y}^2+2hxy+2gx+2fy+c=0,$
the condition of perpendicularity $|a+b|=0$ can be applied.

$\Rightarrow 1+\alpha =0$ or $\alpha =-1$
also line satisfies
$abc+2hgf-a{f}^2-c{h}^2-b{g}^2=0$

so, $a^2-{\beta }^2=0$

$\beta =a.$