If frac{x^2}{a} + frac{y^2}{b} + frac{2xy}{h} | vedclass

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

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$9 : 8$

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(A) The given equation is $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$. Dividing by $x^2$,we get $\frac{1}{a} + \frac{2}{h}(\frac{y}{x}) + \frac{1}{b}(\frac{y}{x})^2 = 0$. Let $m = \frac{y}{x}$ be the slope of the lines. Then $\frac{1}{b}m^2 + \frac{2}{h}m + \frac{1}{a} = 0$. Let the slopes be $m_1$ and $m_2$. Given $m_2 = 2m_1$. From the quadratic equation,sum of roots $m_1 + m_2 = -\frac{2/h}{1/b} = -\frac{2b}{h}$. So,$3m_1 = -\frac{2b}{h} \Rightarrow m_1 = -\frac{2b}{3h}$. Product of roots $m_1 m_2 = \frac{1/a}{1/b} = \frac{b}{a}$. So,$2m_1^2 = \frac{b}{a} \Rightarrow 2(-\frac{2b}{3h})^2 = \frac{b}{a}$. $2(\frac{4b^2}{9h^2}) = \frac{b}{a} \Rightarrow \frac{8b^2}{9h^2} = \frac{b}{a}$. $\frac{b}{h^2} = \frac{9}{8a} \Rightarrow \frac{ab}{h^2} = \frac{9}{8}$. Thus,$ab : h^2 = 9 : 8$.

If $\frac{x^2}{a} + \frac{y^2}{b} + \frac{2xy}{h} = 0$ represents a pair of straight lines and the slope of one line is twice the other,then $ab : h^2$ is

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