Answer
$y=\sqrt 3 x+2$ and $-8$
Work Step by Step
Here, we have $x'=2 \pi \cos 2 \pi t$ and $y'=-2 \pi \sin 2 \pi t$
This implies that $\dfrac{dy}{dx}=-\tan (2 \pi t)$
Also, we have $x=\dfrac{-\sqrt 3}{2}$ and $y=\dfrac{1}{2}$
$\dfrac{dy}{dx}(t=\dfrac{-1}{6})=-\tan (2 \pi) (\dfrac{-1}{6})=-1$
Now, $y-(\dfrac{1}{2})=(\sqrt 3) (x-(\dfrac{-\sqrt 3}{2}))$
Thus, $y= \sqrt 3 x+2$
Now, $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{dy'}{dt}}{\dfrac{dx}{dt}}=-\sec^3 2 \pi t$
$\dfrac{d^2y}{dx^2}(t=\dfrac{-1}{6})=-\sec^3 2 \pi (\dfrac{-1}{6})=\dfrac{-1}{\cos^3 (\dfrac{\pi}{3})}$
Hence, $\dfrac{d^2y}{dx^2}=\dfrac{-1}{(\dfrac{1}{2})^3}=-8$
