Question

What is the equation of the line normal to `f(x)=lnx^2-1/x^2` at `x=-2`?

Answer 1

The normal line to this function at the given point has equation `y=4/5 (x+2)+ln(4)-1/4=4/5 x+ln(4)+27/20`

Explanation:

First note that `f(-2)=ln(4)-1/4`, so the graph of the normal line goes through the point `(-2,ln(4)-1/4)`.

Second, `f'(x)=2/x+2/x^3` so that `f'(-2)=-1-1/4=-5/4` is the slope of the tangent line to the graph of `f` at the given point.

Since the normal line has slope equal to the negative reciprocal of the tangent line (since they are perpendicular), the normal line has slope `-1/(-5/4)=4/5`.

Putting all this together gives the equation of the normal line to be `y=4/5 (x+2)+ln(4)-1/4=4/5 x+ln(4)+27/20`.