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

Feb 12, 2017

The normal line to this function at the given point has equation $y = \frac{4}{5} \left(x + 2\right) + \ln \left(4\right) - \frac{1}{4} = \frac{4}{5} x + \ln \left(4\right) + \frac{27}{20}$

#### Explanation:

First note that $f \left(- 2\right) = \ln \left(4\right) - \frac{1}{4}$, so the graph of the normal line goes through the point $\left(- 2 , \ln \left(4\right) - \frac{1}{4}\right)$.

Second, $f ' \left(x\right) = \frac{2}{x} + \frac{2}{x} ^ 3$ so that $f ' \left(- 2\right) = - 1 - \frac{1}{4} = - \frac{5}{4}$ 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 $- \frac{1}{- \frac{5}{4}} = \frac{4}{5}$.

Putting all this together gives the equation of the normal line to be $y = \frac{4}{5} \left(x + 2\right) + \ln \left(4\right) - \frac{1}{4} = \frac{4}{5} x + \ln \left(4\right) + \frac{27}{20}$.