Question

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

Answer 1

`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.