Asked by Tony
How to convert y^2-8x-16=0 to polar form
Answers
Answered by
Reiny
y = rsinØ, x = rcosØ
y^2 - 8x - 16 = 0
r^2 sin^2 Ø - 8rcosØ - 16 = 0
y^2 - 8x - 16 = 0
r^2 sin^2 Ø - 8rcosØ - 16 = 0
Answered by
Angelica
Can somone check my work please
y = rsinØ, x = rcosØ
y^2 - 8x - 16 = 0
r^2 sin^2 Ø - 8rcosØ - 16 = 0
(rsin(Ø))^2 - 8rcos(Ø) = 16
r^2sin^2(Ø) = 16+8rcos(Ø)
divide by r on both sides:
rsin^2(Ø) = 16+8cos(Ø)
divide by sin^2(Ø) to get r by itself:
r = 16+8cos(Ø) / (sin^2(Ø))
r = 16+8cos(Ø)csc^2(Ø)
y = rsinØ, x = rcosØ
y^2 - 8x - 16 = 0
r^2 sin^2 Ø - 8rcosØ - 16 = 0
(rsin(Ø))^2 - 8rcos(Ø) = 16
r^2sin^2(Ø) = 16+8rcos(Ø)
divide by r on both sides:
rsin^2(Ø) = 16+8cos(Ø)
divide by sin^2(Ø) to get r by itself:
r = 16+8cos(Ø) / (sin^2(Ø))
r = 16+8cos(Ø)csc^2(Ø)
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