Find the equation of the tangent to the curve | vedclass

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(B) The given curve is $y = -\frac{2}{x - 3}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = \frac{2}{(x - 3)^2}$.Given that the sl

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$y - 2x + 10 = 0$

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(B) The given curve is $y = -\frac{2}{x - 3}$.Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = \frac{2}{(x - 3)^2}$.Given that the slope of the tangent is $2$,we set $\frac{2}{(x - 3)^2} = 2$.This simplifies to $(x - 3)^2 = 1$,which means $x - 3 = \pm 1$.Thus,$x = 4$ or $x = 2$.For $x = 4$,$y = -\frac{2}{4 - 3} = -2$. The point is $(4, -2)$.For $x = 2$,$y = -\frac{2}{2 - 3} = 2$. The point is $(2, 2)$.The equation of the tangent at $(4, -2)$ with slope $2$ is $y - (-2) = 2(x - 4)$,which simplifies to $y + 2 = 2x - 8$ or $y - 2x + 10 = 0$.The equation of the tangent at $(2, 2)$ with slope $2$ is $y - 2 = 2(x - 2)$,which simplifies to $y - 2 = 2x - 4$ or $y - 2x + 2 = 0$.Comparing with the given options,$y - 2x + 10 = 0$ is the correct choice.

Find the equation of the tangent to the curve $y + \frac{2}{x - 3} = 0$ which has a slope of $2$.

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