(A) The given equation is $x^2+4xy+4y^2+3x+6y-4=0$. This can be written as $(x+2y)^2+3(x+2y)-4=0$. Let $t = x+2y$,then $t^2+3t-4=0$. $(t+4)(t-1)=0$,so

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(A) The given equation is $x^2+4xy+4y^2+3x+6y-4=0$. This can be written as $(x+2y)^2+3(x+2y)-4=0$. Let $t = x+2y$,then $t^2+3t-4=0$. $(t+4)(t-1)=0$,so $(x+2y+4)(x+2y-1)=0$. The two parallel lines are $L_1: x+2y+4=0$ and $L_2: x+2y-1=0$. The distance $\lambda$ between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is given by $\lambda = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$. $\lambda = \frac{|4-(-1)|}{\sqrt{1^2+2^2}} = \frac{5}{\sqrt{5}} = \sqrt{5}$. Therefore,$\lambda^2 = (\sqrt{5})^2 = 5$.

Class 11 Mathematics Pair of straight lines Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

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

MHT CET 2023 All MHT CET

A
$5$
B
$\sqrt{5}$
C
$25$
D
$\frac{9}{5}$

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Pair of straight lines

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Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

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