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

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(C) The given equation is $ax^2+2hxy-2ay^2+3x+15y-9=0$. Since the lines intersect at $(1,1)$,the point $(1,1)$ must satisfy the equation: $a(1)^2+2h(1

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-$7$

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(C) The given equation is $ax^2+2hxy-2ay^2+3x+15y-9=0$. Since the lines intersect at $(1,1)$,the point $(1,1)$ must satisfy the equation: $a(1)^2+2h(1)(1)-2a(1)^2+3(1)+15(1)-9=0$ $a+2h-2a+3+15-9=0$ $-a+2h+9=0 \implies a-2h=9$ (Equation $1$). For a general second-degree equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ to represent a pair of lines,the condition is $ABC+2FGH-AF^2-BG^2-CH^2=0$. Here,$A=a, H=h, B=-2a, G=3/2, F=15/2, C=-9$. Substituting these values: $a(-2a)(-9) + 2(15/2)(h)(3/2) - a(15/2)^2 - (-2a)(3/2)^2 - (-9)(h)^2 = 0$ $18a^2 + \frac{45h}{2} - \frac{225a}{4} + \frac{18a}{4} + 9h^2 = 0$ $18a^2 + 9h^2 + \frac{45h}{2} - \frac{207a}{4} = 0$ Using $a=2h+9$ from Equation $1$,substituting into the condition and solving for $h$ yields $h=-3$ and $a=3$. Thus,$ah = 3 	imes (-3) = -9$. Wait,re-evaluating the intersection point condition: the partial derivatives with respect to $x$ and $y$ must be zero at $(1,1)$. $f_x = 2ax+2hy+3 = 0 \implies 2a+2h+3=0$ $f_y = 2hx-4ay+15 = 0 \implies 2h-4a+15=0$ Solving these: $2a+2h=-3$ and $-4a+2h=-15$. Subtracting: $6a=12 \implies a=2$. Then $2(2)+2h=-3 \implies 2h=-7 \implies h=-3.5$. $ah = 2 	imes (-3.5) = -7$.

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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