Polar Equation Help Please!

[Homer J]

--- question: convert x^2 + 4x + y^2 - 2y = 0 into a polar equation

thanks

[samiira]

x^2 + 4x + y^2 -2y = 0


Know that the conversion formulas that relate the point (x,y) and (r, θ) are: x= rcos θ, y=rsin θ, r²= x² + y² and tan θ= y/x. These are important for any type of conversion between the two forms as well as some trigonometric identities

(x^2 + y^2) +4x - 2y = 0

subsitute r^2 = x^2 + y^2 aswell as y= rsin(t)  ,  x= rcos(t)

so becomes

r^2 - 2rsin(t) +4rcos(t) = 0

factorise

r [r -2sin(t) + 4cos(t) ] = 0

r= 0 and (r -2sin(t) + 4cos(t)) = 0

The equation R = 0 is the pole. But the pole is included in the graph of the second equation r - 2sin(t) + 4cos(t) = 0 .
We therefore can keep only the second equation.

now solve for r to get polar equation

r - 2 sin(t) + 4cos(t) = 0

r = 2 sin(t) - 4 cos(t)


correct me if i am wrong

[TrueTears]

converting cartesian to polar is normally done in a systematic way, substitute x = rcos(theta), y = rsin(theta), substitute the in the cartesan and solve for r.

convince yourself why we have the relationship x = f(theta,r) and y = f(theta, r)