Question

The combined equation of two lines $L$ and $L_1$ is $2a^2+axy+3y^2=0$ and the combined equation of two lines $L$ and $L_2$ is $2x^2+bxy-3y^2=0$. If $L_1$ and $L_2$ are perpendicular, then $a^2+b^2=$

• A. 26
• B. 29
• C. 13
• D. 85

Answer 1

Answer: (A)

Let $L\Rightarrow y=mx$,
$L_1\Rightarrow y=kx$
$L_2\Rightarrow y=-\frac{1}{k}x$
Now, $(y-mx)(y-kx)=0$
$y^2-ykx-mxy+mk{x}^2=0$
$\Rightarrow mk{x}^2-(k+m)xy+y^2=0$
Given that, $2x^2+axy+3y^2=0$
or $\frac{2}{3}{x}^2+\frac{a}{3}xy+y^2=0$
On comparing, $mk=\frac{2}{3^{\prime }}-(k+m)=\frac{a}{3}\dots$(i)
Now, $(y-mx)\left(y+\frac{x}{k}\right)=0$
$\Rightarrow y^2+\frac{xy}{k}-mxy-\frac{m{x}^2}{k}=0$
$\Rightarrow -\frac{m}{k}{x}^2+\left(\frac{1}{k}-m\right)xy+y^2=0$
which is $2x^2+bxy-3y^2=0$
or $-\frac{2}{3}{x}^2-\frac{b}{3}xy+y^2=0$
On comparing, $\frac{-m}{k}=\frac{-2}{3},-\frac{b}{3}=\frac{1}{k}-m$
So, $m=\frac{2k}{3},-\frac{b}{3}=\frac{1-km}{k}\dots$(ii)
By solving Eqs. (i) and (ii), we get
$m=\frac{2}{3},k=1$ and $a=-5,b=-1$
$\therefore a^2+b^2=25+1=26$