Find an equation of the tangent line at the given point on the curve

Given the parametric equations, find an equation of the tangent line at the given point on the curve.





1. The problem statement, all variables and given/known data



Find an equation of the tangent line at each given point on the curve:





x = 2cotΘ and y=2sin[itex]^{2}θ[/itex] at point ([itex]\frac{-2}{\sqrt{3}}[/itex],[itex]\frac{3}{2})[/itex]







2. Relevant equations



dy/dx = dy/dθ/dx/dθ



3. The attempt at a solution



[itex]\frac{-2}{\sqrt{3}} = 2cotθ[/itex] and [itex]\frac{3}{2}=2sin^{2}θ[/itex]



[itex]\frac{-1}{\sqrt{3}}=\frac{cosθ}{sinθ



}[/itex] and [itex]\frac{\sqrt{3}}{2}=sinθ[/itex]



letting θ = [itex]\frac{2\pi}{3}[/itex] for each function yields the point ([itex]\frac{-2}{\sqrt{3}}[/itex],[itex]\frac{3}{2})[/itex]





dy/dx = [itex]\frac{dy/dθ}{dx/dθ}[/itex] = [itex](\frac{2sin^{2}θ}{2cotθ})'[/itex]



= [itex]-2sin^{3}θcosθ[/itex]



at θ= [itex]\frac{2\pi}{3}[/itex]



slope of tangent = 9/8



so m = (y-y)/(x-x)



9/8 = [itex]\frac{(y-\frac{3}{2}}{x-(-\frac{-2}{\sqrt{3}}}[/itex]



y = [itex]\frac{9}{8} + 15/4[/itex]





Answer doesn't match. I must have made a mistake somewhere.