How do you convert #3y= 2x^2-2x-y^2 # into a polar equation?

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Answer 1 · 2018-07-26T01:25:04

#r = 0# and
#r = ( 2 cos theta + 3 sin theta)/(2 cos^2 theta - sin^2 theta ) #.
This hyperbola passes through the pole #( O ), r = 0#.

Conversion formula:

#( x. y ) = r ( cos theta, sin theta ), r = sqrt (x^2 + y^2) >= 0#

#2x^2 - y^2- 2x - 3y = 0# converts to

#r = 0# and

#r(2 cos^2 theta - sin^2 theta ) - ( 2 cos theta + 3 sin theta)) = 0#.

This hyperbola passes through the pole #( O ), r = 0#.

See graph.

graph{2x^2 - y^2- 2x - 3y = 0 [-20 20 -12 8] }

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