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

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Answer 1 · 2016-02-21T07:00:26

#theta=(3pi)/4=135^@# and #theta=tan^-1 (-1/2)=153.435^@#

Start with the given

#3xy=-x^2-2y^2#

#x^2+3xy+2y^2=0#

do factoring to simplify

#(x+y)(x+2y)=0#

Use #x=r cos theta# and #y=r sin theta#

#(x+y)(x+2y)=0#

#(r cos theta+r sin theta)(r cos theta+2*r sin theta)=0#

cancel all the #r#s

#(cos theta+ sin theta)( cos theta+2* sin theta)=0#

equate both factors to zero

#cos theta+ sin theta=0#

#sin theta=-cos theta#

#tan theta=-1#

#theta=(3pi)/4=135^@#

For the other factor:

#cos theta+2* sin theta=0#

#2*sin theta=-cos theta#

#tan theta=-1/2#

#theta=tan^-1 (-1/2)=153.435^@#

These are 2 lines passing thru the Origin (0, 0) with
slopes =-1 and -1/2

graph{3xy=-x^2-2y^2[-20,20,-10,10]}

have a nice day !