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

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(B) Given the curve $xy = 1$,we have $y = \frac{1}{x}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{x^2}$.The slope of

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$a > 0, b < 0$

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(B) Given the curve $xy = 1$,we have $y = \frac{1}{x}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{x^2}$.The slope of the tangent at any point $(x, y)$ is $-\frac{1}{x^2}$.The slope of the normal is the negative reciprocal of the tangent slope,which is $x^2$.The slope of the line $ax + by + c = 0$ is $-\frac{a}{b}$.Since the line is a normal to the curve,we equate the slopes: $x^2 = -\frac{a}{b}$.Since $x^2$ is always non-negative for real $x$,we must have $-\frac{a}{b} > 0$,which implies $\frac{a}{b} This condition holds if $a$ and $b$ 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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