## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=2 \cos t \\
y=\sin t
\end{array}\right. \\
& t_{0}=-\frac{\pi}{3}
\end{aligned}
$$
## Solution

Since $t_{0}=-\frac{\pi}{3}$, then

$x_{0}=2 \cos \left(-\frac{\pi}{3}\right)=2 \cdot 0.5=1$

$y_{0}=\sin \left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}$

Let's find the derivatives:

$x_{t}^{\prime}=(2 \cos t)^{\prime}=-2 \sin t$

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

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{\cos t}{-2 \sin t}=-\frac{1}{2 \tan t}$

Then:

$y_{0}^{\prime}=-\frac{1}{2 \tan t}=-\frac{1}{2 \cdot(-\sqrt{3})}=\frac{\sqrt{3}}{6}$

Equation of the tangent line:

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

$y-\left(-\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{6} \cdot(x-1)$

$y=\frac{\sqrt{3}}{6} \cdot(x-1)-\frac{\sqrt{3}}{2}$

$y=\frac{\sqrt{3}}{6} \cdot x-\frac{\sqrt{3}}{6}-\frac{\sqrt{3}}{2}$

$y=\frac{\sqrt{3}}{6} \cdot x-\frac{2 \sqrt{3}}{3}$

Equation of the normal line:

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

$y-\left(-\frac{\sqrt{3}}{2}\right)=-\frac{1}{\left(\frac{\sqrt{3}}{6}\right)}(x-1)$

$y=-2 \sqrt{3}(x-1)-\frac{\sqrt{3}}{2}$

$y=-2 \sqrt{3} \cdot x+2 \sqrt{3}-\frac{\sqrt{3}}{2}$

$y=-2 \sqrt{3} \cdot x+\frac{3 \sqrt{3}}{2}$

Kuznetsov Differentiation 17-26