# How do you find parametric equations of a tangent line?

Feb 15, 2017

The parametric equations of the tangent line to the curve $y = f \left(x\right)$ in the point $\left({x}_{0} , f \left({x}_{0}\right)\right)$ are:

$\left\{\begin{matrix}x = {x}_{0} + t \\ y = f \left({x}_{0}\right) + f ' \left({x}_{0}\right) t\end{matrix}\right.$

#### Explanation:

Given a curve $y = f \left(x\right)$, the slope intercept form of the equation of the tangent line to the point $\left({x}_{0} , f \left({x}_{0}\right)\right)$ is:

$y \left(x\right) = f \left({x}_{0}\right) + f ' \left({x}_{0}\right) \left(x - {x}_{0}\right)$

So, if we pose:

$x = {x}_{0} + t$

we have:

$y = f \left({x}_{0}\right) + f ' \left({x}_{0}\right) \left({x}_{0} + t - {x}_{0}\right) = f \left({x}_{0}\right) + f ' \left({x}_{0}\right) t$

The parametric equations are then:

$\left\{\begin{matrix}x = {x}_{0} + t \\ y = f \left({x}_{0}\right) + f ' \left({x}_{0}\right) t\end{matrix}\right.$