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