How do you convert #3x^2+6xy-y^2=9# into polar form?

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Answer 1 · 2016-02-16T07:53:24

#r=3/sqrt(3*cos^2 theta+6*sin theta*cos theta-sin^2 theta)#

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

from the given

#3x^2+6xy-y^2=9#

#3(r cos theta)^2+6(r cos theta)(r sin theta)-(r sin theta)^2=9#

#3r^2 cos^2 theta+6r^2*sin theta*cos theta-r^2 sin^2 theta=9#

#r^2(3 cos^2 theta+6*sin theta*cos theta- sin^2 theta)=9#

#r^2=9/(3 cos^2 theta+6*sin theta*cos theta- sin^2 theta)#

#r=sqrt(9/(3 cos^2 theta+6*sin theta*cos theta- sin^2 theta))#

#r=3/sqrt(3 cos^2 theta+6*sin theta*cos theta- sin^2 theta)#

have a nice day... from the Philippines