Before we get started, let's remember the essential equation for converting between Cartesian and polar coordinates:
x = rcos(theta), y = rsin(theta) x=rcos(θ),y=rsin(θ)
So once we have x and y, we can substitute in. Hence,
2 = (8y-x)^2 - 5y - x2=(8y−x)2−5y−x
2 = 64y^2 - 16xy + x^2 - 5y - x 2=64y2−16xy+x2−5y−x
2 = 64r^2sin^2theta - 16r^2 sin theta cos theta + r^2 cos^2 theta - 5r sin theta - r cos theta 2=64r2sin2θ−16r2sinθcosθ+r2cos2θ−5rsinθ−rcosθ
2=r^2(64sin^2 theta + cos^2 theta - 16 cos theta sin theta) - r (5sin theta + cos theta) 2=r2(64sin2θ+cos2θ−16cosθsinθ)−r(5sinθ+cosθ)
This is kinda the end if we want to have an implicit answer. However, if we want to have a formula for rr in terms of thetaθ we need to realize that the above is a quadratic. So that the formula will fit on one line, let c = cos(theta)c=cos(θ) and s = sin(theta)s=sin(θ):
r = ((5s+c) pm sqrt((5s+c)^2 + 4 cdot 2 cdot (64 s^2 + c^2 - 16sc)))/(2(64s^2 + c^2 - 16 sc)) r=(5s+c)±√(5s+c)2+4⋅2⋅(64s2+c2−16sc)2(64s2+c2−16sc)
r = ((5s+c) pm sqrt((5s+c)^2 + 8(8s-c)^2))/(2(8s-c)^2) r=(5s+c)±√(5s+c)2+8(8s−c)22(8s−c)2
yielding two curves.
