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(C) The given equation is $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$. Multiplying by $abh$, we get $bhx^2 + 2axy + ahy^2 = 0$.Dividing by $x^

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$9:8$

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(C) The given equation is $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$. Multiplying by $abh$, we get $bhx^2 + 2axy + ahy^2 = 0$.Dividing by $x^2$, we get $ahy^2/x^2 + 2ay/x + bh = 0$. Let $m = y/x$ be the slope of the lines.Then $ahm^2 + 2am + bh = 0$.Let the roots be $m$ and $2m$. Sum of roots: $m + 2m = 3m = -2a / ah = -2/h$, so $m = -2/(3h)$.Product of roots: $m \cdot 2m = 2m^2 = bh / ah = b/a$.Substituting $m = -2/(3h)$ into $2m^2 = b/a$: $2(-2/(3h))^2 = b/a \implies 2(4 / 9h^2) = b/a \implies 8 / 9h^2 = b/a$.Rearranging gives $h^2 / ab = 9 / 8$, so $h^2 : ab = 9:8$.

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

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