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

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Aicharya · Accepted Answer

Let $L \Rightarrow y=m x, L_1 \Rightarrow y=k x, L_2 \Rightarrow y=-\frac{1}{k} x$ Now, $ \begin{aligned} \text { Now, } & (y-m x)(y-k x)=0 \\ & y^2-y k x-m x y+m k x^2=0 \\ \Rightarrow \quad & m k x^2-(k+m) x y+y^2=0 \end{aligned} $ Given that, $2 x^2+a x y+3 y^2=0$ or $\quad \frac{2}{3} x^2+\frac{a}{3} x y+y^2=0$

Now, $ (y-m x)\left(y+\frac{x}{k}\right)=0 $ $ \begin{array}{rlrl} & \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 \end{array} $ which is or $ \begin{aligned} 2 x^2+b x y-3 y^2 & =0 \\ -\frac{2}{3} x^2-\frac{b}{3} x y+y^2 & =0 \end{aligned} $ On comparing, $\frac{-m}{k}=\frac{-2}{3},-\frac{b}{3}=\frac{1}{k}-m$

By solving Eqs. (i) and (ii), we get $ \begin{aligned} m & =\frac{2}{3}, k=1 \text { and } a=-5, b=-1 \\ \therefore \quad a^2+b^2 & =25+1=26 \end{aligned} $

The combined equation of two lines $L$ and $L_1$ is $2 x^2+a x y+3 y^2=0$ and the combined equation of two…

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