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

Question

1722 views around the world

You can reuse this answer
Creative Commons License

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.