Answer
$y=\dfrac{\sqrt 3}{2}x+\dfrac{1}{4}$ and $\dfrac{d^2y}{dx^2}=\dfrac{1}{4}$
Work Step by Step
Take the derivative of the given equations and isolate the variables.
$ \dfrac{dx}{dt}=\dfrac{1}{2}\sec^2 t$
and $\dfrac{dy}{dt}=\dfrac{1}{2}\sec t \tan t$
Now, Slope: $\dfrac{dy}{dx}=\sin t$
At $t=\dfrac{\pi}{3}$
Then $ \dfrac{dy}{dx}=\sin(\dfrac{\pi}{3})=\dfrac{\sqrt 3}{2}$
Now, $y-1=(\dfrac{\sqrt 3}{2})(x-\dfrac{\sqrt 3}{2})\implies y=\dfrac{\sqrt 3}{2}x+\dfrac{1}{4}$
Now, $\dfrac{d^2y}{dx^2}=\dfrac{dy'/dt}{dx/dt}=2 \cos^3 t$
At $t=\dfrac{\pi}{3}$
$\dfrac{d^2y}{dx^2}=2 \cos^3 (\dfrac{\pi}{3})=\dfrac{1}{4}$
