Horizontal and Vertical Tangents for Polar Equation

[EnlightenedOne]

Homework Statement

Find the points at which the following polar equation has horizontal and vertical tangents:

r² = 4cos(2θ)

Homework Equations

[itex]\frac{dy}{dx}[/itex] = [itex]\frac{r'(θ)sinθ + r(θ)cosθ}{r'(θ)cosθ - r(θ)sinθ}[/itex]

Horizontal Tangent: [itex]\frac{dy}{dθ}[/itex] = 0; [itex]\frac{dx}{dθ}[/itex] ≠ 0
Vertical Tangent: [itex]\frac{dx}{dθ}[/itex] = 0; [itex]\frac{dy}{dθ}[/itex] ≠ 0

The Attempt at a Solution

There is no "clean" way of solving for r (because of the +/- sqrt) so that I could find r'(θ) to use in the formula. So, I figured I would use implicit differentiation:

r² = 4cos(2θ)

2r[itex]\frac{dr}{dθ}[/itex] = -8sin(2θ)

[itex]\frac{dr}{dθ}[/itex] = [itex]\frac{-4sin(2θ)}{r}[/itex]

But, when I plug it in the formula, I now have an r and a θ when I set the numerator and denominator of [itex]\frac{dy}{dx}[/itex] equal to zero (separately of course). I don't know what to do when I have both variables like that and I'm trying to solve for θ. How should I be approaching this problem? Have I done this problem right so far? If so, what do I do next? If not, any suggestions? Please be clear. I can't find this problem answered clearly anywhere on the internet.
Thank you

 

[EnlightenedOne]

For some reason the text isn't showing up right on my computer after posting. Let me know if its the same for you so I can edit and try to fix it.

 

[haruspex]

But, when I plug it in the formula, I now have an r and a θ when I set the numerator and denominator of [itex]\frac{dy}{dx}[/itex] equal to zero (separately of course). I don't know what to do when I have both variables like that and I'm trying to solve for θ.

You should have an equation involving r and θ (no derivatives around). Your original equation for the curve also has those two variables. Two equations, two unknowns.
If still stuck, please post all your working.