The second-degree equation x^2 + 2sqrt{2}xy + | vedclass

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(C) The given equation is $x^2 + 2\sqrt{2}xy + 2y^2 + 4x + 4\sqrt{2}y + 1 = 0$.Comparing this with the general second-degree equation $ax^2 + 2hxy + b

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$2$

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(C) The given equation is $x^2 + 2\sqrt{2}xy + 2y^2 + 4x + 4\sqrt{2}y + 1 = 0$.Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a=1, h=\sqrt{2}, b=2, g=2, f=2\sqrt{2}, c=1$.Since $h^2 - ab = (\sqrt{2})^2 - (1)(2) = 2 - 2 = 0$,the lines are parallel.The distance between two parallel lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by $d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}}$.Substituting the values: $d = 2\sqrt{\frac{2^2 - (1)(1)}{1(1 + 2)}} = 2\sqrt{\frac{4 - 1}{3}} = 2\sqrt{\frac{3}{3}} = 2\sqrt{1} = 2$.

The second-degree equation $x^2 + 2\sqrt{2}xy + 2y^2 + 4x + 4\sqrt{2}y + 1 = 0$ represents a pair of parallel straight lines. The distance between them is

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