How do you convert #2y= -x^2+4xy # into a polar equation?

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Answer 1 · 2018-01-05T08:26:21

#r = (2sin(theta))/ (4cos(theta)sin(theta)-cos^2(theta))#

Given #2y=-x^2 + 4xy#

and

#x=rcos(theta)# and #y=rsin(theta)# we can substitute:

#2rsin(theta) = -(rcos(theta))^2+4(rcos(theta))(rsin(theta))#

#2rsin(theta) = -r^2cos^2(theta)+4r^2cos(theta)sin(theta)#

Since #r ne 0# for all #theta# we can divide through by #r#:

#2sin(theta) = -rcos^2(theta)+4rcos(theta)sin(theta)#

Now we can factor #r# from the right side:

#2sin(theta) = r(-cos^2(theta)+4cos(theta)sin(theta))#

Now we can solve for #r#:

#r = (2sin(theta))/ (-cos^2(theta)+4cos(theta)sin(theta))#

and for no very good reason, I'd rather rewrite the denominator:

#r = (2sin(theta))/ (4cos(theta)sin(theta)-cos^2(theta))#