## problem statement

Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.

$\left\{\begin{array}{l}x=\frac{t+1}{t} \\ y=\frac{t-1}{t}\end{array}\right.$

$t_{0}=-1$
## Solution

Since $t_{0}=-1$, then

$x_{0}=\frac{-1+1}{-1}=0$

$y_{0}=\frac{-1-1}{-1}=2$

Find the derivatives:

$x_{t}^{\prime}=\left(\frac{t+1}{t}\right)^{\prime}=\frac{1 \cdot t-(t+1) \cdot 1}{t^{2}}=\frac{t-t-1}{t^{2}}=-\frac{1}{t^{2}}$

$y_{t}^{\prime}=\left(\frac{t-1}{t}\right)^{\prime}=\frac{1 \cdot t-(t-1) \cdot 1}{t^{2}}=\frac{t-t+1}{t^{2}}=\frac{1}{t^{2}}$

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{1}{t^{2}}\right) /\left(\frac{1}{t^{2}}\right)=-1$

Then

$y_{0}^{\prime}=-1$

Equation of the tangent line:

$$
\begin{aligned}
& y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right) \\
& y-2=-1 \cdot(x-0) \\
& y=-x+2
\end{aligned}
$$

Equation of the normal line:

$y-y_{0}=-\frac{1}{y_{0}^{\prime}}\left(x-x_{0}\right)$

$y-2=-\frac{1}{-1} \cdot(x-0)$

$y=x+2$

## Problem Kuznetsov Differentiation 17-18