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

Answers (1)

We know that the slope of the tangent of at the point of the given curve is given by  \frac{dy}{dx}

Given the equation of curve is
y = \frac{1}{x-3}
\frac{dy}{dx} = \frac{-1}{(x-3)^2}
It is given that slope is 2
So,
\frac{-1}{(x-3)^2} = 2 \Rightarrow (x-3)^2 = \frac{-1}{2} = x-3 = \pm \frac{\sqrt-1}{\sqrt2} \\ \\
So, this is not possible as our coordinates are imaginary numbers
Hence, no tangent is possible with slope 2  to the curve  y = \frac{1}{x-3}

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