The distance between the pair of lines x^2+2 | vedclass

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(C) The given equation of the pair of lines is $x^2+2 \sqrt{2} x y+2 y^2+4 x+4 \sqrt{2} y+1=0$.This can be written as $(x+\sqrt{2}y)^2 + 4(x+\sqrt{2}y

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The given equation of the pair of lines is $x^2+2 \sqrt{2} x y+2 y^2+4 x+4 \sqrt{2} y+1=0$.This can be written as $(x+\sqrt{2}y)^2 + 4(x+\sqrt{2}y) + 1 = 0$.Let $X = x+\sqrt{2}y$. Then the equation becomes $X^2 + 4X + 1 = 0$.Solving for $X$ using the quadratic formula $X = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $X = \frac{-4 \pm \sqrt{16-4}}{2} = -2 \pm \sqrt{3}$.So,the two parallel lines are $x+\sqrt{2}y + 2 - \sqrt{3} = 0$ and $x+\sqrt{2}y + 2 + \sqrt{3} = 0$.The distance $d$ between two parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.Here,$A=1, B=\sqrt{2}, C_1=2-\sqrt{3}, C_2=2+\sqrt{3}$.$d = \frac{|(2-\sqrt{3})-(2+\sqrt{3})|}{\sqrt{1^2+(\sqrt{2})^2}} = \frac{|-2\sqrt{3}|}{\sqrt{1+2}} = \frac{2\sqrt{3}}{\sqrt{3}} = 2$.

The distance between the pair of lines $x^2+2 \sqrt{2} x y+2 y^2+4 x+4 \sqrt{2} y+1=0$ is:

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