If the line ax + by + c = 0 is a normal to th | vedclass

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(C) Given the curve $xy = 1$,differentiating with respect to $x$ gives $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.The slope of the no

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$a  0$ or $a > 0, b < 0$

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(C) Given the curve $xy = 1$,differentiating with respect to $x$ gives $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.The slope of the normal at any point $(x, y)$ on the curve is $m_n = -\frac{1}{dy/dx} = \frac{x}{y}$.The given line is $ax + by + c = 0$,which can be written as $y = -\frac{a}{b}x - \frac{c}{b}$. The slope of this line is $m = -\frac{a}{b}$.Equating the slopes,we get $-\frac{a}{b} = \frac{x}{y}$,which implies $\frac{a}{b} = -\frac{x}{y}$.Since $xy = 1$,either both $x, y > 0$ or both $x, y If $x, y > 0$,then $\frac{x}{y} > 0$,so $\frac{a}{b}  0, b  0$.If $x, y  0$,so $\frac{a}{b}  0, b  0$.Thus,the condition is $a  0$ or $a > 0, b < 0$.

If the line $ax + by + c = 0$ is a normal to the curve $xy = 1$,then:

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