Question

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

Answer 1

Please see the explanation for the process.

`r = -((3cos(theta))/(3(cos(theta))(sin(theta)) - 2))`

Explanation:

Substitute `rcos(theta)` for x and `rsin(theta)` for y:

`3(rcos(theta))(rsin(theta)) = 2(rcos(theta))^2 - 3rcos(theta) + 2(rsin(theta))^2`

Factor out r whenever possible:

`3r^2(cos(theta))(sin(theta)) = 2r^2cos^2(theta) - 3rcos(theta) + 2r^2sin^2(theta)`

Swap the last two terms:

`3r^2(cos(theta))(sin(theta)) = 2r^2cos^2(theta) + 2r^2sin^2(theta) - 3rcos(theta)`

Combine the `r^2` terms on the right:

`3r^2(cos(theta))(sin(theta)) = 2r^2(cos^2(theta) + sin^2(theta)) - 3rcos(theta)`

Substitute 1 for `cos^2(theta) + sin^2(theta):`

`3r^2(cos(theta))(sin(theta)) = 2r^2 - 3rcos(theta)`

Combine like terms:

`r^2(3cos(theta))(sin(theta)) - 2) = - 3rcos(theta)`

Move everything to the left side:

`r^2(3cos(theta))(sin(theta)) - 2) + 3rcos(theta) = 0`

Divide both sides by `(3cos(theta))(sin(theta)) - 2)`

`r^2 + r((3cos(theta))/(3(cos(theta))(sin(theta)) - 2)) = 0`

Divide both sides by r:

`r + ((3cos(theta))/(3(cos(theta))(sin(theta)) - 2)) = 0`

Solve for r:

`r = -((3cos(theta))/(3(cos(theta))(sin(theta)) - 2))`