If the distance between the parallel lines given by the equation $x^2+4 x y+4 y^2+3 x+6y-4=0$ is $\lambda$, then $\lambda^2=$

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Aicharya · Accepted Answer

Given equation is $\begin{aligned} & x^2+4 x y+4 y^2+3 x+6 y-4=0 \ & (x+2 y)^2+3(x+2 y)-4=0 \ & (x+2 y+4)(x+2 y-1)=0 \end{aligned}$ $	herefore \quad$ The lines are: $x+2 y+4=0$ and $x+2 y-1=0$ $\begin{aligned} 	herefore \quad 	ext { Required distance } & =\frac{|4-(-1)|}{\sqrt{1+4}} \ & =\frac{5}{\sqrt{5}}=\sqrt{5} 	ext { units } \end{aligned} $ $\lambda=\sqrt{5}$ units $\lambda^2=5$

If the distance between the parallel lines given by the equation $x^2+4 x y+4 y^2+3 x+6y-4=0$ is $\lambda$,…

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