How do you convert #3xy=-x^2+4y^2 # into a polar equation?

1 Answer
May 13, 2016

#cos(theta)sin(theta)+cos(theta)^2+4sin(theta)^2=0#

Explanation:

We must be carefull with those equations. Here #4y^2-x^2-3x y = 0# can be written as #(4y-x)(x+y) = 0# so, the function graphic is build with two straigths: #4y-x=0# and #x + y=0#. This phenomenon will appear as well in polar coordinates. Making #x = r cos(theta), y = r sin(theta)# and substituting, we get:
#r^2 cos(theta)sin(theta) = -r^2 cos(theta)^2+4 r^2 sin(theta)^2# and after simplification, we get the condition: #cos(theta)sin(theta)+cos(theta)^2+4sin(theta)^2=0#. Solving for #theta# we will obtain the straigths directions.