Line $l$ is tangent to the curve $y=2-(1/x)$ at the point where $x=p$. Show that an equation of $l$ may be expressed in the form $p^2y-x= 2p^2-2p$.

Question

$x$ not equal to $0$. A little confused on how to proceed. I can get as far as taking the derivative to find the slope of the tangent line: $dy/dx=1/x^2$. Not specifically after the answer, just a hint on what to do next.

Answer 1

The line has to pass through the point:

$$x_0=p,\quad y_0=2-\frac1{x_0}=2-\frac 1p\tag{1}$$

The slope of the line is defined by the first derivative:

$$y'=\frac{d}{dx}(2-\frac 1x)=\frac1{x^2}$$

At point $(x_0, y_0)$ the value of derivative is:

$$y'_0=\frac1{p^2}\tag{2}$$

Equation of the line passing through the point $(x_0,y_0)$ with slope $y'_0$ is:

$$y-y_0=y'_0 (x-x_0)\tag{3}$$

Replace (1) and (2) into (3) and you are done.