How do you convert $r = 16 / 3 -5 cos theta$ into cartesian form?

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Answer 1 · 2017-01-30T18:01:20

Please see the explanation.

Here is the graph of $r = 16/3 - 5cos(theta)$

![Desmos.com]( useruploads.socratic.org )

Multiply both sides of the equation by r:

$r^2 = 16/3r-5rcos(theta)$

Subsitute $x^2+y^2" for "r^2$ :

$x^2+y^2 = 16/3r-5rcos(theta)$

Substitute $sqrt(x^2+y^2)$ for r:

$x^2+y^2 = 16/3sqrt(x^2+y^2)-5rcos(theta)$

Substitute x for $rcos(theta)$

$(x^2+y^2) = 16/3sqrt(x^2+y^2)-5x$

Here is a graph of the converted equation:

![Desmos.com]( useruploads.socratic.org )

Please observe that the graphs are identical.

How do you convert r = 16 / 3 -5 cos thetar = 16 / 3 -5 cos theta into cartesian form?