How do you convert #r=6/(2-3sin(theta))# into cartesian form?

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Answer 1 · 2016-08-15T06:27:29

#4x^2-5y^2+9y-9=0#

Use #r(cos theta, sin theta)=(x, y)#.

#r = sqrt(x^2+y^2) and sin theta = y/r#

Now the given equation becomes

#r = (6r)/(2-(3x)/r)#. Canceling r and rearranging,

#r=3-3/2y#. Squaring, ,

#x^2+y^2=9/4y^2-9y+9#. Simplifying.

#4x^2-5y^2+9y-9=0#