The equation 2x^2 + 4xy - py^2 + 4x + qy + 1 | vedclass

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(C) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. Comparing this with $2x^2 + 4xy - py^2 + 4x + qy + 1 =

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$p = 2$ and $q = 0$ or $8$

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(C) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. Comparing this with $2x^2 + 4xy - py^2 + 4x + qy + 1 = 0$,we get $a = 2, h = 2, b = -p, g = 2, f = q/2, c = 1$. For the lines to be perpendicular,the condition is $a + b = 0$. $2 + (-p) = 0 \Rightarrow p = 2$. For the equation to represent a pair of straight lines,the determinant condition is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. Substituting the values: $2(-p)(1) + 2(q/2)(2)(2) - 2(q/2)^2 - (-p)(2)^2 - 1(2)^2 = 0$. With $p = 2$: $2(-2)(1) + 4q - 2(q^2/4) - (-2)(4) - 4 = 0$. $-4 + 4q - q^2/2 + 8 - 4 = 0$. $-q^2/2 + 4q = 0 \Rightarrow q^2 - 8q = 0$. $q(q - 8) = 0 \Rightarrow q = 0$ or $q = 8$. Thus,$p = 2$ and $q = 0$ or $8$.

The equation $2x^2 + 4xy - py^2 + 4x + qy + 1 = 0$ will represent two mutually perpendicular straight lines,if

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