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(D) The equation of the pair of lines is $ax^2+2hxy+by^2=0$. Let the slopes be $m$ and $m^2$.From the properties of the quadratic equation in $y/x$,we

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$6$

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(D) The equation of the pair of lines is $ax^2+2hxy+by^2=0$. Let the slopes be $m$ and $m^2$.From the properties of the quadratic equation in $y/x$,we have:Sum of slopes: $m+m^2 = -\frac{2h}{b} \quad (1)$Product of slopes: $m \cdot m^2 = m^3 = \frac{a}{b} \quad (2)$From $(1)$,$m(1+m) = -\frac{2h}{b}$. Cubing both sides:$m^3(1+m)^3 = -\frac{8h^3}{b^3}$$m^3(1+m^3+3m(1+m)) = -\frac{8h^3}{b^3}$Substitute $m^3 = \frac{a}{b}$ and $m(1+m) = -\frac{2h}{b}$:$\frac{a}{b} \left(1 + \frac{a}{b} + 3\left(-\frac{2h}{b}ight)ight) = -\frac{8h^3}{b^3}$$\frac{a}{b} \left(\frac{b+a-6h}{b}ight) = -\frac{8h^3}{b^3}$$a(a+b-6h) = -8h^3$$a^2+ab-6ah = -8h^3$Dividing by $abh$ (assuming $a, b, h 
eq 0$):$\frac{a}{bh} + \frac{1}{h} - \frac{6}{b} = -\frac{8h^2}{ab}$Rearranging terms gives $\frac{a+b}{h} + \frac{8h^2}{ab} = 6$.

If the slope of one of the lines represented by $ax^2+2hxy+by^2=0$ is the square of the other,then $\frac{a+b}{h}+\frac{8h^2}{ab}$ is equal to:

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