Condition of the problem

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

$\left\{\begin{array}{l}x=3 \cos t \\ y=4 \sin t\end{array}\right.$

$t_{0}=\frac{\pi}{4}$
## Solution

Since $t_{0}=\frac{\pi}{4}$, then

$x_{0}=3 \cos \frac{\pi}{4}=\frac{3 \sqrt{2}}{2}$

$y_{0}=4 \sin \frac{\pi}{4}=\frac{4 \sqrt{2}}{2}=2 \sqrt{2}$

Let's find the derivatives:

$x_{t}^{\prime}=(3 \cos t)^{\prime}=-3 \sin t$

$y_{t}^{\prime}=(4 \sin t)^{\prime}=4 \cos t$

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{4 \cos t}{-3 \sin t}=-\frac{4}{3} \cdot \operatorname{ctg} t$

Then:

$y_{0}^{\prime}=-\frac{4}{3} \cdot \operatorname{ctg} \frac{\pi}{4}=-\frac{4}{3} \cdot 1=-\frac{4}{3}$

Equation of the tangent line:

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

$$
\begin{aligned}
& y-2 \sqrt{2}=-\frac{4}{3} \cdot\left(x-\frac{3 \sqrt{2}}{2}\right) \\
& y=-\frac{4}{3} \cdot x+2 \sqrt{2}+2 \sqrt{2} \\
& y=-\frac{4}{3} \cdot x+4 \sqrt{2}
\end{aligned}
$$

Equation of the normal line:

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

$y-2 \sqrt{2}=-\frac{1}{\left(-\frac{4}{3}\right)} \cdot\left(x-\frac{3 \sqrt{2}}{2}\right)$

$y=\frac{3}{4} \cdot\left(x-\frac{3 \sqrt{2}}{2}\right)+2 \sqrt{2}$

$y=\frac{3}{4} \cdot x-\frac{9 \sqrt{2}}{8}+\frac{16 \sqrt{2}}{8}$

$y=\frac{3}{4} \cdot x+\frac{7 \sqrt{2}}{8}$

## Problem Kuznetsov Differentiation 17-23