Question

If one of the lines given by the equation $2x^2+pxy+3y^2=0,$ coincides with one of those given by $2x^2+qxy-3y^2=0$ and the other lines represented by them are perpendicular, then $(p,q)$ can be

• A. $(-5,1)$
• B. $(5,1)$
• C. $(5,-1)$
• D. $(-5,-1)$

Answer 1

Answer: (B), (D)

The correct options are
B $(5,1)$
D $(-5,-1)$

Let $2x^2+pxy+3y^2=(y-mx)(y-m^{\prime }x)$
and $2x^2+qxy-3y^2=(y+\frac{1}{m}x)(y-m^{\prime }x)$
Then, $m+m^{\prime }=\frac{-p}{3},m{m}^{\prime }=\frac{2}{3}$
and $\frac{1}{m}-m^{\prime }=\frac{-q}{3},-\frac{m^{\prime }}{m}=-\frac{2}{3}$
$\Rightarrow \frac{m{m}^{\prime }}{-m^{\prime }/m}=\frac{2/3}{-2/3}$
$\Rightarrow m^2=1\Rightarrow m=\pm 1$
If $m=1,$
then $m^{\prime }=\frac{2}{3}$ and so $p=-5,q=-1.$
If $m=-1,$
then $m^{\prime }=-\frac{2}{3}$ and so $p=5,q=1.$