Answer
$y=\sqrt 3 x+2$ and $\dfrac{d^2y}{dx^2}=-8$
Work Step by Step
Here, $x'=2 \pi \cos 2 \pi t; y'=-2 \pi \sin 2 \pi t$
and $\dfrac{dy}{dx}=-\tan (2 \pi t)$
Now, at $t=\dfrac{-1}{6}$
we have $x=\dfrac{-\sqrt 3}{2}$ and $y=\dfrac{1}{2}$
$\dfrac{dy}{dx}=-\tan (2 \pi) (\dfrac{-1}{6})=-1$
Now, $y-(\dfrac{1}{2})=(\sqrt 3) (x-(\dfrac{-\sqrt 3}{2})) \implies y= \sqrt 3 x+2$
$\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{dy'}{dt}}{\dfrac{dx}{dt}}=-\sec^3 2 \pi t$
Now, at $t=\dfrac{\pi}{4}$
$\dfrac{d^2y}{dx^2}=-\sec^3 2 \pi (\dfrac{-1}{6})$
or, $\dfrac{-1}{\cos^3 (\dfrac{\pi}{3})}=\dfrac{-1}{(\dfrac{1}{2})^3}$
Thus, $\dfrac{d^2y}{dx^2}=-8$
