Answer
Tangent line: $\quad y=\sqrt{3}x$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\frac{2\pi}{3}}=0$
Work Step by Step
The point $P$ on the curve corresponding to $t=\displaystyle \frac{2\pi}{3}$ is
$\displaystyle \left\{\begin{array}{l}
x=\cos\frac{2\pi}{3}=-\frac{1}{2}\\
\\
y=\sqrt{3}\cos\frac{2\pi}{3}=-\frac{\sqrt{3}}{2}
\end{array}\right.\quad, P(-\frac{1}{2},-\frac{\sqrt{3}}{2})$
The slope of the tangent line at $t=\displaystyle \frac{2\pi}{3}$ is $\left. \displaystyle \frac{dy}{dx} \right|_{t=\frac{2\pi}{3}}$
$\displaystyle \left\{\begin{array}{l}
\frac{dx}{dt}=-\sin t,\\
\\
\frac{dy}{dt}=-\sqrt{3}\sin t
\end{array}\right.\quad, \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-\sqrt{3}\sin t}{-\sin t}=\sqrt{3}$
$\left. \displaystyle \frac{dy}{dx} \right|_{t=\frac{2\pi}{3}}=\sqrt{3}$
Use the point-slope equation for the tangent line.
$y-(-\displaystyle \frac{\sqrt{3}}{2})=\sqrt{3}(x-(-\frac{1}{2}))$
$y=\displaystyle \sqrt{3}x+\frac{\sqrt{3}}{2} -\frac{\sqrt{3}}{2}$
$y=\sqrt{3}x$
With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\frac{2\pi}{3}}$,
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{\frac{d}{dt}[\sqrt{3}]}{-\sin t}=0$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\frac{2\pi}{3}}=0$
