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(D) The given equation is $a x^2+2 h x y+b y^2=0$. Let the slopes of the lines be $m$ and $m^2$. From the properties of the homogeneous equation,the s

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$6$

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(D) The given equation is $a x^2+2 h x y+b y^2=0$. Let the slopes of the lines be $m$ and $m^2$. From the properties of the homogeneous equation,the sum of slopes is $m+m^2 = -\frac{2h}{b}$ and the product of slopes is $m \cdot m^2 = m^3 = \frac{a}{b}$. We have $m(1+m) = -\frac{2h}{b}$. Cubing both sides,we get $m^3(1+m)^3 = -\frac{8h^3}{b^3}$. Expanding this,$m^3(1+m^3+3m(1+m)) = -\frac{8h^3}{b^3}$. Substituting $m^3 = \frac{a}{b}$ and $m(1+m) = -\frac{2h}{b}$,we get $\frac{a}{b}(1+\frac{a}{b}+3(-\frac{2h}{b})) = -\frac{8h^3}{b^3}$. Multiplying by $b^3$,we get $a(b+a-6h) = -8h^3$. $ab + a^2 - 6ah = -8h^3$. Dividing by $abh$,we get $\frac{a+b}{h} + \frac{8h^2}{ab} = 6$.

If the slope of one of the lines represented by $a x^2+2 h x y+b y^2=0$ is the square of the other,then $\frac{a+b}{h}+\frac{8 h^2}{a b}$ is

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