How would you convert this particular polar equation to cartesian?

Question

How would you go about converting the polar equation $r^2 = 4cos(2\theta)$ into a cartesian equation in terms of y?

I have just started working on polar-cartesian equations, but do not yet have sufficient pattern recognition. My current approach of trying to apply the basic trigonometric pythagorean identity is leading me back to where I started.

Any hints on how I should approach this?

Answer 1

if you multiply the equation by $r^2,$ you get $$r^4 = 4r^2\cos 2\theta = 4r^2(\cos^2 \theta - \sin^2 \theta) = 4(x^2 - y^2) $$ that is $$(x^2 + y^2 )^2 = 4x^2 - 4y^2 \to y^4+2y^2(x^2+2) +(x^4-4x^2).$$ use the quadratic formula to get $$y^2 = -x^2 - 2\pm\sqrt{(x^2+2)^2 -x^4 +4x^2}=-2-x^2 \pm 2\sqrt{2x^2+1} $$ finally, $$y = \pm\sqrt{2\sqrt{2x^2 + 1} - x^2 - 2}. $$

Answer 2

Hint: $x^2+y^2=r^2=4\cos (2\theta) = 4\cos^2\theta - 4\sin^2 \theta = \dfrac{4x^2-4y^2}{x^2+y^2}$