Find the equation of all lines having slope 2 | vedclass

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(D) The given curve is $y = -\frac{2}{x-3}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{2}{(x-3)^2}$.Since the slope of the tang

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$y-2x+2=0$ and $y-2x+10=0$

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(D) The given curve is $y = -\frac{2}{x-3}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{2}{(x-3)^2}$.Since the slope of the tangent is given as $2$,we have $\frac{2}{(x-3)^2} = 2$.This implies $(x-3)^2 = 1$,so $x-3 = \pm 1$.Thus,$x = 4$ or $x = 2$.If $x = 4$,then $y = -\frac{2}{4-3} = -2$. The point is $(4, -2)$.The equation of the tangent at $(4, -2)$ is $y - (-2) = 2(x - 4)$,which simplifies to $y + 2 = 2x - 8$ or $y - 2x + 10 = 0$.If $x = 2$,then $y = -\frac{2}{2-3} = 2$. The point is $(2, 2)$.The equation of the tangent at $(2, 2)$ is $y - 2 = 2(x - 2)$,which simplifies to $y - 2 = 2x - 4$ or $y - 2x + 2 = 0$.Therefore,the equations of the tangents are $y - 2x + 2 = 0$ and $y - 2x + 10 = 0$.

Find the equation of all lines having slope $2$ and being tangent to the curve $y+\frac{2}{x-3}=0.$

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