Rectangular to Polar Equation

[dulaney11]

Could someone please check my work?

y^2 - 8x - 16 = 0

r^2-x^2-8x-16=0
r^2=(x+4)
r= square root of (x+4) is that as simplified as it can be?

 

[royhaas]

Your substitution should be \(\displaystyle x = r\cos(\theta), y = r\sin(\theta)\).

 

[fasteddie65]

Could someone please check my work?

y^2 - 8x - 16 = 0

r^2-x^2-8x-16=0
r^2=(x+4)
r= square root of (x+4) is that as simplified as it can be?

This can't be right, because the equation still has an "x" in it.

y^2 - 8x - 16 = 0
y^2 = 8x + 16
(r sin ø)^2 = 8(r cos ø) + 16
r^2 sin^2 ø = 8r cos ø + 16
r^2 sin^2 ø - 8r cos ø - 16 = 0
This is a quadratic equation in r. You can use the Quadratic Formula to find r in terms of ø now.

r = {-(-8 cos ø) ± ?[(-8 cos ø)^2 - 4(sin^2 ø)(-16)]}/4(sin^2 ø) = [8 cos ø ± ?(64 cos^2 ø + 64 sin^2 ø)]/4 sin^2 ø = (8 cos ø ± 8)/4 (1 - cos^2 ø) = (2 cos ø ± 2)/[(1 - cos ø)(1 + cos ø)] = (2 cos ø + 2)/[(1 - cos ø)(1 + cos ø)] = 2/(1 - cos ø) or -2/(1 + cos ø)

Actually, this is one of the easier conversions. Most conversions with conic sections turn out to be very complicated. And don't get me started with conversions of polar equations like lemniscates or limacons to rectangular form.