If the line ax + by + c = 0, ab neq 0, is a t | vedclass

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(B) Given the curve $xy = 1 - 2x$. Rewriting the equation as $y = \frac{1 - 2x}{x} = \frac{1}{x} - 2$. Differentiating with respect to $x$, we get $\f

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$a > 0, b > 0$

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(B) Given the curve $xy = 1 - 2x$. Rewriting the equation as $y = \frac{1 - 2x}{x} = \frac{1}{x} - 2$. Differentiating with respect to $x$, we get $\frac{dy}{dx} = -\frac{1}{x^2}$. Since $\frac{dy}{dx} The line $ax + by + c = 0$ can be written as $y = -\frac{a}{b}x - \frac{c}{b}$. The slope of this line is $m = -\frac{a}{b}$. Since the line is tangent to the curve, its slope must be equal to the derivative at the point of tangency, which is negative. Therefore, $-\frac{a}{b} 0$. This condition $\frac{a}{b} > 0$ holds if both $a$ and $b$ have the same sign. Thus, either $a > 0, b > 0$ or $a < 0, b < 0$.

If the line $ax + by + c = 0, ab \neq 0,$ is a tangent to the curve $xy = 1 - 2x,$ then

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