If ax^2+2hxy-2ay^2+3x+15y-9=0 represents a pa | vedclass

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(C) The general equation of a pair of lines is $ax^2+2hxy+by^2+2gx+2fy+c=0$. Given the equation $ax^2+2hxy-2ay^2+3x+15y-9=0$,we have $b = -2a$,$2g = 3

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-$7$

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(C) The general equation of a pair of lines is $ax^2+2hxy+by^2+2gx+2fy+c=0$. Given the equation $ax^2+2hxy-2ay^2+3x+15y-9=0$,we have $b = -2a$,$2g = 3$,$2f = 15$,and $c = -9$.Since the lines intersect at $(x_0, y_0) = (1, 1)$,the partial derivatives with respect to $x$ and $y$ must be zero at this point.$\frac{\partial}{\partial x} (ax^2+2hxy-2ay^2+3x+15y-9) = 2ax+2hy+3$. At $(1, 1)$,$2a+2h+3 = 0 \implies 2a+2h = -3$.$\frac{\partial}{\partial y} (ax^2+2hxy-2ay^2+3x+15y-9) = 2hx-4ay+15$. At $(1, 1)$,$2h-4a+15 = 0 \implies 2h-4a = -15$.Subtracting the two equations: $(2a+2h) - (2h-4a) = -3 - (-15) \implies 6a = 12 \implies a = 2$.Substituting $a=2$ into $2a+2h = -3$: $2(2)+2h = -3 \implies 4+2h = -3 \implies 2h = -7 \implies h = -3.5$.Thus,$ah = 2 	imes (-3.5) = -7$.

If $ax^2+2hxy-2ay^2+3x+15y-9=0$ represents a pair of lines intersecting at $(1,1)$,then $ah=$

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