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 $L \Rightarrow y=m x$,
$ L_{1} \Rightarrow y=k x$
$ L_{2} \Rightarrow y=-\frac{1}{k} x$
Now, $(y-m x)(y-k x)=0$
$y^{2}-y k x-m x y+m k x^{2}=0$
$\Rightarrow m k x^{2}-(k+m) x y+y^{2}=0$
Given that, $2 x^{2}+a x y+3 y^{2}=0$
or $\frac{2}{3} x^{2}+\frac{a}{3} x y+y^{2}=0$
On comparing, $m k=\frac{2}{3'}-(k+m)=\frac{a}{3} \dots$(i)
Now, $(y-m x)\left(y+\frac{x}{k}\right)=0$
$\Rightarrow y^{2}+\frac{x y}{k}-m x y-\frac{m x^{2}}{k}=0$
$\Rightarrow -\frac{m}{k} x^{2}+\left(\frac{1}{k}-m\right) x y+y^{2}=0$
which is $2 x^{2}+b x y-3 y^{2}=0$
or $-\frac{2}{3} x^{2}-\frac{b}{3} x y+y^{2}=0$
On comparing, $\frac{-m}{k}=\frac{-2}{3},-\frac{b}{3}=\frac{1}{k}-m$
So, $m=\frac{2 k}{3},-\frac{b}{3}=\frac{1-k m}{k} \dots$(ii)
By solving Eqs. (i) and (ii), we get
$m =\frac{2}{3}, k=1 $ and $ a=-5, b=-1$
$\therefore a^{2}+b^{2} =25+1=26$