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(B) Given $\alpha$ and $\beta$ are roots of $x^2+ax-b=0$. From Vieta's formulas, $\alpha+\beta = -a$ and $\alpha\beta = -b$. Differentiating the curve

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$\frac{4}{c^2}$

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(B) Given $\alpha$ and $\beta$ are roots of $x^2+ax-b=0$. From Vieta's formulas, $\alpha+\beta = -a$ and $\alpha\beta = -b$. Differentiating the curve $(\frac{x}{\alpha})^n + (\frac{y}{\beta})^n = 2$ with respect to $x$: $\frac{n}{\alpha}(\frac{x}{\alpha})^{n-1} + \frac{n}{\beta}(\frac{y}{\beta})^{n-1} \frac{dy}{dx} = 0$. At the point $(\alpha, \beta)$, the slope of the tangent is $\frac{dy}{dx} = -\frac{\beta}{\alpha}$. The line $x \cos 	heta + y \sin 	heta = c$ has a slope of $-\cot 	heta$. Equating the slopes: $-\cot 	heta = -\frac{\beta}{\alpha} \implies \cot 	heta = \frac{\beta}{\alpha}$. Since the line passes through $(\alpha, \beta)$, $\alpha \cos 	heta + \beta \sin 	heta = c$. Using $\cot 	heta = \frac{\beta}{\alpha}$, we have $\cos 	heta = \frac{\beta}{\sqrt{\alpha^2+\beta^2}}$ and $\sin 	heta = \frac{\alpha}{\sqrt{\alpha^2+\beta^2}}$. Substituting these into the line equation: $\alpha(\frac{\beta}{\sqrt{\alpha^2+\beta^2}}) + \beta(\frac{\alpha}{\sqrt{\alpha^2+\beta^2}}) = c \implies \frac{2\alpha\beta}{\sqrt{\alpha^2+\beta^2}} = c$. We need to evaluate $(\frac{a}{b})^2 + \frac{2}{b} = (\frac{-(\alpha+\beta)}{-\alpha\beta})^2 + \frac{2}{-\alpha\beta} = \frac{(\alpha+\beta)^2}{(\alpha\beta)^2} - \frac{2}{\alpha\beta} = \frac{\alpha^2+\beta^2+2\alpha\beta-2\alpha\beta}{(\alpha\beta)^2} = \frac{\alpha^2+\beta^2}{(\alpha\beta)^2}$. From $\frac{2\alpha\beta}{\sqrt{\alpha^2+\beta^2}} = c$, we get $\alpha^2+\beta^2 = \frac{4\alpha^2\beta^2}{c^2}$. Substituting this into the expression: $\frac{4\alpha^2\beta^2/c^2}{(\alpha\beta)^2} = \frac{4}{c^2}$.

Let $\alpha, \beta$ be two roots of the quadratic equation $x^2+ax-b=0, b \neq 0$. If the straight line $x \cos \theta + y \sin \theta = c$ touches the curve $(\frac{x}{\alpha})^n + (\frac{y}{\beta})^n = 2$ at the point $(\alpha, \beta)$, then $(\frac{a}{b})^2 + \frac{2}{b} =$

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