Question

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

Answer 1

`theta=(3pi)/4=135^@` and `theta=tan^-1 (-1/2)=153.435^@`

Explanation:

Start with the given

`3xy=-x^2-2y^2`

`x^2+3xy+2y^2=0`

do factoring to simplify

`(x+y)(x+2y)=0`

Use `x=r cos theta` and `y=r sin theta`

`(x+y)(x+2y)=0`

`(r cos theta+r sin theta)(r cos theta+2*r sin theta)=0`

cancel all the `r`s

`(cos theta+ sin theta)( cos theta+2* sin theta)=0`

equate both factors to zero

`cos theta+ sin theta=0`

`sin theta=-cos theta`

`tan theta=-1`

`theta=(3pi)/4=135^@`

For the other factor:

`cos theta+2* sin theta=0`

`2*sin theta=-cos theta`

`tan theta=-1/2`

`theta=tan^-1 (-1/2)=153.435^@`

These are 2 lines passing thru the Origin (0, 0) with
slopes =-1 and -1/2

graph{3xy=-x^2-2y^2[-20,20,-10,10]}

have a nice day !