# How do you convert 8=(4x-11y)^2+3y into polar form?

Jun 18, 2017

Substitute $x = r \cos \left(\theta\right) \mathmr{and} y = r \sin \left(\theta\right)$.

#### Explanation:

Given $x = r \cos \left(\theta\right) \mathmr{and} y = r \sin \left(\theta\right)$, we have
$8 = {\left(4 r \cos \theta - 11 r \sin \theta\right)}^{2} + 3 r \sin \theta$

$8 = {\left(r \left[4 \cos \theta - 11 \sin \theta\right]\right)}^{2} + 3 r \sin \theta$

$8 = {r}^{2} {\left(4 \cos \theta - 11 \sin \theta\right)}^{2} + 3 r \sin \theta$

We can use FOIL to simplify the expression in parentheses, but it doesn't lead us anywhere "nice."

We could solve for r using the Quadratic Formula if we are required to, but this is positively unfriendly, and the equation is already in a Polar form.