If the equation 12x^2 + 7xy - py^2 - 18x + qy | vedclass

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(A) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if $abc + 2fgh - af^2 -

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$p = 12, q = 1$

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(A) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. Comparing the given equation $12x^2 + 7xy - py^2 - 18x + qy + 6 = 0$ with the general form,we have $a = 12, h = 7/2, b = -p, g = -9, f = q/2, c = 6$. Since the lines are perpendicular,the coefficient of $x^2$ plus the coefficient of $y^2$ must be zero,so $a + b = 0$ $\Rightarrow 12 - p = 0$ $\Rightarrow p = 12$. Substituting $p = 12$ into the condition for a pair of lines: $12(-p)(6) + 2(q/2)(-9)(7/2) - 12(q/2)^2 - (-p)(-9)^2 - 6(7/2)^2 = 0$. Substituting $p = 12$: $-864 - 63q/2 - 3q^2 + 12(81) - 6(49/4) = 0$. $-864 - 31.5q - 3q^2 + 972 - 73.5 = 0$. $-3q^2 - 31.5q + 34.5 = 0$. Dividing by $-1.5$: $2q^2 + 21q - 23 = 0$. $(2q + 23)(q - 1) = 0$. Thus,$q = 1$ (taking the integer solution).

If the equation $12x^2 + 7xy - py^2 - 18x + qy + 6 = 0$ represents a pair of perpendicular straight lines,then

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