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

How do you convert 3x^2+6xy-y^2=93x^2+6xy-y^2=9 into polar form?