How do you convert 8=(4x-11y)^2+3y8=(4x−11y)2+3y into polar form?

1 Answer
namralos
Jun 18, 2017

Substitute x = rcos(theta) and y = rsin(theta)x=rcos(θ)andy=rsin(θ).

Explanation:

Given x = rcos(theta) and y = rsin(theta)x=rcos(θ)andy=rsin(θ), we have
8 = (4rcostheta - 11rsintheta)^2 + 3rsintheta8=(4rcosθ−11rsinθ)2+3rsinθ

8 = (r[4costheta - 11sintheta])^2 + 3rsintheta8=(r[4cosθ−11sinθ])2+3rsinθ

8 = r^2(4costheta - 11sintheta)^2 + 3rsintheta8=r2(4cosθ−11sinθ)2+3rsinθ

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.

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