Question

If one of the lines given by the equation $2x^2+axy+3y^2=0$ coincide with one of those given by $2x^2+bxy-3y^2=0$ and the other lines represented by them be perpendicular, then

• A. $a=-5,b=1$
• B. $a=5,b=-1$
• C. $a=5,b=1$
• D. none of these

Answer 1

The correct option is C $a=5,b=1$

Let $\frac{2}{3}{x}^2+\frac{a}{3}xy+y^2=(y-mx)(y-m^{\prime }x)$
and $\frac{2}{-3}{x}^2+\frac{b}{-3}xy+y^2=(y+\frac{1}{m}x)(y-m^{\prime }x)$
then $m+m^{\prime }=-\frac{a}{3},m{m}^{\prime }=\frac{2}{3}\left(i\right)$
$\frac{1}{m}-m^{\prime }=\frac{-b}{3},-\frac{m^{\prime }}{m}=-\frac{2}{3}\left(ii\right)$
$\Rightarrow m^2=1\Rightarrow m=\pm 1$
If $m=1,m^{\prime }=\frac{2}{3}\Rightarrow a=-5,b=-1$
If $m=-1,m^{\prime }=-\frac{2}{3}\Rightarrow a=5,b=1$