How do you convert #6=(5x+4y)^2-7x# into polar form?

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Answer 1 · 2018-07-16T04:44:10

#r^2(4\cos \theta + 5\sin \theta)^2-7r\cos \theta-6=0#.

Just remember these coordinate conversions:

#x=r\cos \theta#

#y=r\sin \theta#

#r=\sqrt{x^2+y^2}#

So the given equation is converted to

#6=(4r\cos \theta + 5r\sin \theta)^2-7r\cos \theta#.

We can rearrange this to:

#r^2(4\cos \theta + 5\sin \theta)^2-7r\cos \theta-6=0#.