Question

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

Answer 1

The curve is $y+\frac{2}{x-3}=0$

Slope of the tangent to the curve at point $(x,y)$ is $\frac{dy}{dx}$

$\frac{dy}{dx}+\frac{d}{dx}\left(\frac{2}{x-3}\right)=0$

$\Rightarrow \frac{dy}{dx}=-\frac{d}{dx}\left(\frac{2}{x-3}\right)$

$\Rightarrow \frac{dy}{dx}=-\left(\frac{-2}{{(x-3)}^2}\right)$

Given that slope$=2$

hence,$\frac{dy}{dx}=2$

$\Rightarrow \left(\frac{2}{{(x-3)}^2}\right)=2$

$\Rightarrow {(x-3)}^2=1$

$\Rightarrow x-3=\pm 1$

$\therefore x-3=1,x-3=-1$

$\therefore x=4,2$

If $x=2\Rightarrow y=\frac{-2}{x-3}=\frac{-2}{2-3}=2$

Thus, the point is $(2,2)$

If $x=4\Rightarrow y=\frac{-2}{x-3}=\frac{-2}{4-3}=-2$

Thus, the point is $(4,-2)$

Thus, there are two tangents to the curve with slope $2$ and passing through points $(2,2)$ and $(4,-2)$

Equation of tangent through $(2,2)$ is

$y-2=2(x-2)$

$y-2x+2=0$

Equation of tangent through $(4,-2)$ is

$y-(-2)=2(x-4)$

$y-2x+10=0$