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

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(B) The equation of the curve is $xy = 1$,which can be written as $y = \frac{1}{x}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac

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$a > 0, b < 0$

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(B) The equation of the curve is $xy = 1$,which can be written as $y = \frac{1}{x}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{x^2}$.The slope of the tangent at any point $(x_1, y_1)$ on the curve is $m_t = -\frac{1}{x_1^2}$.The slope of the normal at $(x_1, y_1)$ is $m_n = -\frac{1}{m_t} = x_1^2$.The equation of the 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 normal to the curve,its slope must equal the slope of the normal: $-\frac{a}{b} = x_1^2$.Since $x_1^2 > 0$ for all $x_1 
eq 0$,we must have $-\frac{a}{b} > 0$,which implies $\frac{a}{b} This means $a$ and $b$ must have opposite signs.Looking at the options,if $a > 0$,then $b  0$ (not explicitly listed as a single option,but $b Given the standard form of such problems,$a$ and $b$ must have opposite signs,which is satisfied by $a > 0, b < 0$.

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

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