## Problem Statement

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

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

$t_{0}=2$
## Solution

Since $t_{0}=2$, then

$x_{0}=\frac{3 a \cdot 2}{1+2^{2}}=\frac{6 a}{5}$

$y_{0}=\frac{3 a \cdot 2^{2}}{1+2^{2}}=\frac{12 a}{5}$

Let's find the derivatives:

$$
\begin{aligned}
& x_{t}^{\prime}=\left(\frac{3 a t}{1+t^{2}}\right)^{\prime}=\frac{3 a\left(1+t^{2}\right)-3 a t \cdot 2 t}{\left(1+t^{2}\right)^{2}}=\frac{3 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}} \\
& y_{t}^{\prime}=\left(\frac{3 a t^{2}}{1+t^{2}}\right)^{\prime}=\frac{6 a t\left(1+t^{2}\right)-3 a t^{2} \cdot 2 t}{\left(1+t^{2}\right)^{2}}=\frac{6 a t}{\left(1+t^{2}\right)^{2}} \\
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{6 a t}{\left(1+t^{2}\right)^{2}}\right) /\left(\frac{3 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}\right)=\frac{2 t}{1-t^{2}}
\end{aligned}
$$

Then

$y_{0}^{\prime}=\frac{2 \cdot 2}{1-2^{2}}=-\frac{4}{3}$

Equation of the tangent line:

$y-\frac{12 a}{5}=-\frac{4}{3}\left(x-\frac{6 a}{5}\right)$

$y=-\frac{4}{3} \cdot x+\frac{8 a}{5}+\frac{12 a}{5}$

$y=-\frac{4}{3} \cdot x-4 a$

Equation of the normal line:

$y-\frac{12 a}{5}=-\frac{1}{\left(-\frac{4}{3}\right)}\left(x-\frac{6 a}{5}\right)$

$y=\frac{3}{4} \cdot x-\frac{9 a}{10}+\frac{12 a}{5}$

$y=\frac{3}{4} \cdot x+\frac{3 a}{2}$

## Problem Kuznetsov Differentiation 17-8