Parametric Equations Tanget lines

[600burger]

I am asked to find the equation of the tanget line to the curve at the givien points. (y -y1 = m(x1-x))

The point is:
(-2/sqrt(3), 3/2)

Parametric Equations are:
where t = theta

x = 2*cot(t)
y = 2*sin^2(t)

How would i find what theta is in this set, inorder to solve dy/dx numericlly to get my slope (m)?

 

[cookiemonster]

For what value of theta does x = -2/sqrt(3) and y = 3/2?

cookiemonster

 

[M98Ranger]

To find the value of theta for the given point, we can plug in the x and y coordinates into the parametric equations. This will give us two equations:

-2/sqrt(3) = 2*cot(t)
3/2 = 2*sin^2(t)

We can then solve for t by rearranging the first equation to get cot(t) = -1/sqrt(3), and using the identity cot(t) = cos(t)/sin(t), we can substitute in -1/sqrt(3) for cot(t) in the second equation. This will give us:

3/2 = 2*(1-cos^2(t))

Simplifying, we get:

3/2 = 2-2cos^2(t)

Rearranging and dividing by 2, we get:

cos^2(t) = 1/4

Taking the square root, we get:

cos(t) = ±1/2

From this, we can determine two possible values for theta: t = π/3 and t = 2π/3. We can plug these values back into the original parametric equations to verify that they do indeed give us the given point (-2/sqrt(3), 3/2).

Now that we have determined the value of theta, we can use it to find the slope of the tangent line at that point. We can do this by taking the derivative of both parametric equations with respect to t and then plugging in the value of theta. This will give us:

dx/dt = -2*csc^2(t)
dy/dt = 4*sin(t)*cos(t)

Substituting in t = π/3 or t = 2π/3, we get:

dx/dt = -2*4/3 = -8/3
dy/dt = 4*√3/2 = 2√3

Finally, we can use the formula for the slope of a tangent line, m = dy/dx, to find the slope at the given point:

m = (dy/dt)/(dx/dt) = (2√3)/(-8/3) = -√3/4

Now that we have the slope, we can use the point-slope form of a line to write the equation of the tangent line:

y - (3/2) = (-√3/4)(