If the equation x^2+2 sqrt{2} xy+2y^2+4x+4 sq | vedclass

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(B) The given equation is $x^2+2 \sqrt{2} xy+2y^2+4x+4 \sqrt{2}y+1=0$. This can be written as $(x+\sqrt{2}y)^2 + 4(x+\sqrt{2}y) + 1 = 0$. Let $t = x+\

Vedclass · Accepted Answer

$2 	ext{ units}$

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(B) The given equation is $x^2+2 \sqrt{2} xy+2y^2+4x+4 \sqrt{2}y+1=0$. This can be written as $(x+\sqrt{2}y)^2 + 4(x+\sqrt{2}y) + 1 = 0$. Let $t = x+\sqrt{2}y$. Then the equation becomes $t^2 + 4t + 1 = 0$. Solving for $t$ using the quadratic formula,$t = \frac{-4 \pm \sqrt{16-4}}{2} = -2 \pm \sqrt{3}$. Thus,the two parallel lines are $x+\sqrt{2}y + 2 - \sqrt{3} = 0$ and $x+\sqrt{2}y + 2 + \sqrt{3} = 0$. The distance 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,$A=1, B=\sqrt{2}, C_1=2-\sqrt{3}, C_2=2+\sqrt{3}$. $d = \frac{|(2-\sqrt{3})-(2+\sqrt{3})|}{\sqrt{1^2+(\sqrt{2})^2}} = \frac{|-2\sqrt{3}|}{\sqrt{3}} = \frac{2\sqrt{3}}{\sqrt{3}} = 2 	ext{ units}$. Therefore,the distance between the lines is $2 	ext{ units}$.

If the equation $x^2+2 \sqrt{2} xy+2y^2+4x+4 \sqrt{2}y+1=0$ represents a pair of parallel straight lines,find the distance between them.

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