Find the equation of the normal at the point $(1,2)$ to the curve $y=x+\frac{1}{x}$.

Question

The answer in the textbook is listed as $x=1$, however I don't understand why this is.

My working:

$$y=x+\frac{1}{x}$$

$$f'(x)=1-x^{-2}$$ $$f'(x)=1-\frac{1}{x^{2}}$$ $$f'(1)=1-\frac{1}{1}=0$$

Therefore the gradient of the tangent to the curve at $x=1$ is $0$.

The gradient of the normal is given by:

$$m_1m_2=-1$$ $$m_2=\frac{-1}{0}$$

As anything divided by $0$ is undefined, how do I then use this to find the the normal to the tangent, to the curve, at $x=1$? Is there a way of analytically working out tangents/normals when they are undefined, or is an alternative approach required?

EDIT: My alternative approach would entail ascertaining the tangent at $(1,2)$, which turns out to be $y=2$ - a horizontal line; which means that the normal would be perpendicular to this and is thus given by $x$-value at this point, i.e. $x=1$.

Is a visual way the only method of producing an answer? Is there a better way of demonstrating this?

Answer 1

Perhaps a picture will clarify the situation.

Answer 2

If there’s a chance that you’ll encounter vertical lines, I think that you’re better off working with equations of the form $ax+by+c=0$ instead of $y=mx+b$. This generally avoids the entire issue of having to deal with zero or undefined slopes.

This approach lends itself quite nicely to the method that you outlined in your edit because a common way to construct an equation of this form is to use the point-normal form $\mathbf n\cdot(\mathbf p-\mathbf p_0)=0$. A line normal to a curve at $\mathbf p_0=(x_0,f(x_0))$ is perpendicular to the tangent to the curve there, and the latter has direction vector $(1,f'(x_0))$, therefore an equation of the normal at $\mathbf p_0$ is $$(x-x_0)+f'(x_0)(y-f(x_0))=0.$$ For your function, $f(1)=2$ and $f'(1)=0$, so an equation of the normal at $(1,2)$ is $x-1=0$.