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

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Question

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

Vedclass · Accepted Answer

$2$ units

Vedclass · Answer

(B) The given equation is $x^2 + 2 \sqrt{2} xy + 2y^2 + 4x + 4 \sqrt{2}y + 1 = 0$. This can be rewritten 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 $d$ 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,$A = 1$,$B = \sqrt{2}$,$C_1 = 2 - \sqrt{3}$,and $C_2 = 2 + \sqrt{3}$. $d = \frac{|(2 - \sqrt{3}) - (2 + \sqrt{3})|}{\sqrt{1^2 + (\sqrt{2})^2}} = \frac{|-2 \sqrt{3}|}{\sqrt{3}} = \frac{2 \sqrt{3}}{\sqrt{3}} = 2$ units. Therefore,the correct option is $B$.

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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