# How do you find the equation of a normal line to a curve at a given point?

Aug 1, 2014

The equation of a normal line will have the form

$y = m x + b$

and its slope will be the negative reciprocal of the curve's derivative at the point. That is to say, take the value of the derivative at the point, divide 1 by it, and then multiply that value by $- 1$. You then solve for $b$ after plugging in the $x$ and $y$ coordinates of the point, as well as $m$.

This is much better illustrated with an example:

Let's say that we are expected to find the equation of a line normal to the curve $f \left(x\right) = {x}^{2}$ at the point $\left(2 , 4\right)$. A normal line is a line perpendicular to the tangent line, so we will take the derivative of $f \left(x\right)$ to find the slope of the tangent line, and then take the negative reciprocal of this slope, to find the slope of the normal line.

$\frac{d}{\mathrm{dx}} f \left(x\right) = 2 x$

$\frac{d}{\mathrm{dx}} f \left(2\right) = 2 \cdot 2 = 4$

The negative reciprocal of $4$ is $- \frac{1}{4}$. We now have a value for $m$:

$y = - \frac{1}{4} x + b$

The last step is to plug in the coordinates of our point and solve for $b$:

$4 = - \frac{1}{4} \cdot 2 + b$

$4 = - \frac{2}{4} + b$

$4 + \frac{1}{2} = b$

$b = \frac{9}{2}$

Now we have everything needed to put our full equation together:

$y = - \frac{1}{4} x + \frac{9}{2}$