If the line ax + by + c = 0 is a tangent to t | vedclass

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(C) The equation of the curve is $xy = 4$,which can be written as $y = \frac{4}{x}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} =

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$a < 0, b < 0$

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(C) The equation of the curve is $xy = 4$,which can be written as $y = \frac{4}{x}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{4}{x^2}$.The equation of the line is $ax + by + c = 0$,which can be rewritten as $y = -\frac{a}{b}x - \frac{c}{b}$.The slope of this line is $m = -\frac{a}{b}$.Since the line is a tangent to the curve,the slope of the tangent at any point $(x, y)$ on the curve must equal the slope of the line:$-\frac{4}{x^2} = -\frac{a}{b} \implies \frac{a}{b} = \frac{4}{x^2}$.Since $x^2 > 0$ for all $x 
eq 0$,it follows that $\frac{a}{b} > 0$.This implies that $a$ and $b$ must have the same sign. Looking at the options,the condition $a  0$.

If the line $ax + by + c = 0$ is a tangent to the curve $xy = 4$,then

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