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(B) The given equation is $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$. Dividing by $x^2$,we get $\frac{1}{b}(\frac{y}{x})^2 + \frac{2}{h}(\fra

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$9:8$

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(B) The given equation is $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$. Dividing by $x^2$,we get $\frac{1}{b}(\frac{y}{x})^2 + \frac{2}{h}(\frac{y}{x}) + \frac{1}{a} = 0$. Let the slopes be $m$ and $2m$. From the sum of roots,$m + 2m = 3m = -\frac{2/h}{1/b} = -\frac{2b}{h}$,so $m = -\frac{2b}{3h}$. From the product of roots,$m \cdot 2m = 2m^2 = \frac{1/a}{1/b} = \frac{b}{a}$. Substituting $m$ in the product equation: $2(-\frac{2b}{3h})^2 = \frac{b}{a}$. $2 \cdot \frac{4b^2}{9h^2} = \frac{b}{a}$. $\frac{8b^2}{9h^2} = \frac{b}{a}$. $\frac{ab}{h^2} = \frac{9}{8}$. Thus,$ab : h^2 = 9:8$.

If the slope of one of the two lines represented by $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$ is twice that of the other,then $ab : h^2 = $

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