The equation 8x^2 + 8xy + 2y^2 + 26x + 13y + | vedclass

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(B) The given equation is $8x^2 + 8xy + 2y^2 + 26x + 13y + 15 = 0$. Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,

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$7/(2\sqrt{5})$

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(B) The given equation is $8x^2 + 8xy + 2y^2 + 26x + 13y + 15 = 0$. Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a = 8, h = 4, b = 2, g = 13, f = 13/2, c = 15$. Since $h^2 - ab = 4^2 - (8)(2) = 16 - 16 = 0$,the lines are parallel. The distance between two parallel lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by $d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}}$. Substituting the values: $d = 2\sqrt{\frac{13^2 - (8)(15)}{8(8 + 2)}} = 2\sqrt{\frac{169 - 120}{8(10)}} = 2\sqrt{\frac{49}{80}} = 2 	imes \frac{7}{4\sqrt{5}} = \frac{7}{2\sqrt{5}}$.

The equation $8x^2 + 8xy + 2y^2 + 26x + 13y + 15 = 0$ represents a pair of parallel straight lines. The distance between them is

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