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

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(B) Given curve is $xy = 1$,so $y = \frac{1}{x}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{y}{x} = -\frac{1}{x^2}$.The slope

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$a > 0, b < 0$

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(B) Given curve is $xy = 1$,so $y = \frac{1}{x}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{y}{x} = -\frac{1}{x^2}$.The slope of the tangent at any point $(x, y)$ is $-\frac{1}{x^2}$.The slope of the normal at any point $(x, y)$ is the negative reciprocal of the tangent slope,which is $x^2$.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 $-\frac{a}{b}$.Since the line is a normal to the curve,its slope must equal the slope of the normal at some point $(x, y)$ on the curve.Thus,$-\frac{a}{b} = x^2$.Since $x^2$ is always non-negative for real $x$,and $x^2$ cannot be zero (as the normal exists at points on the curve),we must have $x^2 > 0$.Therefore,$-\frac{a}{b} > 0$,which implies $\frac{a}{b} This means $a$ and $b$ must have opposite signs,i.e.,$a > 0, b  0$.

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

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