# Parametric formula for the tangent line of the curve $x(t) = \cos(t)$, $y(t) = 1 + \sin(t)$, at $(x,y)=(\frac{\sqrt3}2,\frac32)$

Suppose you have the following parametric equations for t $\in [0,2\pi]$:

$x(t) = \cos(t)$, $y(t) = 1 + \sin(t)$

Give the parametric formula for the tangent line of this curve at $(\frac{\sqrt{3}}{2}, \frac{3}{2})$

First, I found at $t = \frac{\pi}{6}$ the $x$ and $y$ points match up. Then, to get the derivative of the tangent line, I used $\frac{dy}{dx} = \frac{\cos(t)}{-\sin(t))}$ evaluated at $\frac{\pi}{6}$. For the tangent line in terms of $x$, I got $y = -\sqrt{3}x + 3$. Now, I don't understand how I am supposed to get the parametric formula. Should I use the tangent line formula from above, or is there a simpler way?

The correct answer is $(x,y) = (\frac{\sqrt{3} - t}{2}, \frac{\sqrt{3}t+3}{2})$ for t $\in (-\infty, \infty)$.

The equation of the tangent line you found is actually $$y=-\sqrt3x+3$$ which can be rewritten as $$y-\tfrac32=-\sqrt3(x-\tfrac{\sqrt3}2)$$ Thus, using the parameter $$s=x-\tfrac{\sqrt3}2$$ a parametric form of the equation of the tangent line is $$x=s+\tfrac{\sqrt3}2\qquad y=\tfrac32-\sqrt3s$$ which you can surely show is equivalent to the answer you were given.