## Problem Statement

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

\[
\left\{
\begin{array}{l}
x=\frac{1+t^{3}}{t^{2}-1} \\
y=\frac{t}{t^{2}-1}
\end{array}
\right.
\]

$t_{0}=2$
## Solution

Since $t_{0}=2$, then

$x_{0}=\frac{1+2^{3}}{2^{2}-1}=\frac{9}{3}=3$

$y_{0}=\frac{2}{2^{2}-1}=\frac{2}{3}$

Let's find the derivatives:

$x_{t}^{\prime}=\left(\frac{1+t^{3}}{t^{2}-1}\right)^{\prime}=\frac{3 t^{2} \cdot\left(t^{2}-1\right)-\left(1+t^{3}\right) \cdot 2 t}{\left(t^{2}-1\right)^{2}}=\frac{3 t^{4}-3 t^{2}-2 t-2 t^{4}}{\left(t^{2}-1\right)^{2}}=$

$=\frac{t^{4}-3 t^{2}-2 t}{\left(t^{2}-1\right)^{2}}$

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

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

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{t^{2}+1}{\left(t^{2}-1\right)^{2}}\right) /\left(\frac{t^{4}-3 t^{2}-2 t}{\left(t^{2}-1\right)^{2}}\right)=-\frac{t^{2}+1}{t^{4}-3 t^{2}-2 t}$

Then:

$y_{0}^{\prime}=-\frac{2^{2}+1}{2^{4}-3 \cdot 2^{2}-2 \cdot 2}=-\frac{5}{16-12-4}=\infty$

Since $y_{0}^{\prime}=\infty$, the equation of the tangent line is:

$x=x_{0}$

$$
x=3
$$

The equation of the normal line:

$$
\begin{aligned}
& y-\frac{2}{3}=-\frac{1}{\infty} \cdot(x-3) \\
& y=-0 \cdot(x-3)+\frac{2}{3} \\
& y=\frac{2}{3}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 17-22