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

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Answer 1 · 2016-07-10T12:19:01

The solution are lines passing by the origin with declivities.

$theta = arctan(1/2)+2kpi$ with $k = 0,pm1,pm2,pm3$

or

$theta = arctan(1/2)+pi+2kpi$ with $k = 0,pm1,pm2,pm3$

Using the pass equations

$x = r cos(theta)$
$y = r sin(theta)$

we arrive at

$3r^2cos(theta)sin(theta)=2r^2cos^2(theta)-2r^2sin^2(theta)$

or

$3cos(theta)sin(theta)=2cos^2(theta)-2sin^2(theta)$

which is a relationship of the type

$f(theta)=0$

with solutions

$theta = arctan(1/2)+2kpi$ with $k = 0,pm1,pm2,pm3$

or

$theta = arctan(1/2)+pi+2kpi$ with $k = 0,pm1,pm2,pm3$

so the solution are lines passing by the origin with those declivities.

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