How do you convert $2 = {(- x - 7 y)}^{2} - 7 x$ $2=(-x-7y)^2-7x$ into polar form?

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Answer 1 · 2018-04-13T09:17:45

$2 = r^{2} {(cos θ + 7 sin θ)}^{2} - 7 r cos θ$ $2=r^2(costheta+7sintheta)^2-7rcostheta$

We will use:
$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$

$2 = {(- r cos θ - 7 r sin θ)}^{2} - 7 r cos θ$ $2=(-rcostheta-7rsintheta)^2-7rcostheta$

$2 = {(- r)}^{2} {(cos θ + 7 sin θ)}^{2} - 7 r cos θ$ $2=(-r)^2(costheta+7sintheta)^2-7rcostheta$

$2 = r^{2} {(cos θ + 7 sin θ)}^{2} - 7 r cos θ$ $2=r^2(costheta+7sintheta)^2-7rcostheta$

This cannot be simplified further and so has to be left as an implivit equation.

How do you convert 2=(-x-7y)^2-7x2=(−x−7y)2−7x2=(-x-7y)^2-7x into polar form?