Find the equation of a tangent line at each given point

[isukatphysics69]

Homework Statement

in title

Homework Equations

x=2cotθ
y=2sin²θ
dy/dx = 2sin³θcosθ
y-y₁=m(x-x₁)
point = (-2/√3,(3/2)​

The Attempt at a Solution

Have been stuck for hours

I solved for the dy/dx above, now I need to figure out how to get rid of the θ to get my equation in terms of x
so I was thinking sqrt(y/2) = sinθ and (x/2)*sinθ = cosθ

is this a correct approach?

 

[Gene Naden]

Apparently $\theta$ is a parameter: $x=x(\theta), y=y(\theta)$. Is that right? So you can write $sin(\theta)=f(y), cos(\theta)=g(y)$. Then you eliminate $\theta$ and get $x=x(y)$ and $\frac{dx}{dy}=h(y)$.

 

[Orodruin]

There is no need to get rid of $\theta$ in order to figure out the slope. What you do need to do is to figure out what value of $\theta$ you should be using, i.e., which value of $\theta$ corresponds to the given point.

 

[Gene Naden]

@Orodruin is right; I don't know what I was thinking of.

 

[isukatphysics69]

How is the point point = (-2/sqrt(3),(3/2) the same as 2pi/3?