# Find the equation of a tangent line at each given point

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?

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)##.

• isukatphysics69
Orodruin
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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.

• isukatphysics69
@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?