Find the equation of all lines having slope 2 | vedclass

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Question

(A) The equation of the given curve is $y = \frac{1}{x-3}, x 
eq 3$. The slope of the tangent to the curve at any point $(x, y)$ is given by the deri

Vedclass · Accepted Answer

No such tangent exists.

Vedclass · Answer

(A) The equation of the given curve is $y = \frac{1}{x-3}, x 
eq 3$. The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$. $\frac{dy}{dx} = \frac{d}{dx} (x-3)^{-1} = -1(x-3)^{-2} = \frac{-1}{(x-3)^2}$. We are given that the slope of the tangent is $2$. Therefore,we set the derivative equal to $2$: $\frac{-1}{(x-3)^2} = 2$. This implies $(x-3)^2 = -\frac{1}{2}$. Since the square of any real number $(x-3)^2$ must be non-negative,it cannot be equal to a negative value $-\frac{1}{2}$. Thus,there is no real value of $x$ for which the slope of the tangent is $2$. Hence,there are no tangents to the given curve with slope $2$.

Find the equation of all lines having slope $2$ which are tangents to the curve $y = \frac{1}{x-3}, x \neq 3$.

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