How do you convert $3y= 2x^2-2xy-7y^2$ into a polar equation?

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Answer 1 · 2017-03-30T18:13:54

Substitute $rcos(theta)$ for $x$ and $rsin(theta)$ for y.
Write $r$ as a function of $theta$

Given: $3y= 2x^2-2xy-7y^2$

Here is the graph of the Cartesian equation:

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

Substitute $rcos(theta)$ for $x$ and $rsin(theta)$ for y.

#3rsin(theta)= 2(rcos(theta))^2-2(rcos(theta))(rsin(theta))-7(rsin(theta))^2

Write $r$ as a function of $theta$

$3sin(theta)r = (2cos^2(theta) -2cos(theta)sin(theta)-7sin^2(theta))r^2$

Please observe that we can safely divide by r, because that will only eliminate the trivial root $r = 0$ :

$3sin(theta) = (2cos^2(theta) -2cos(theta)sin(theta)-7sin^2(theta))r$

Divide by the coefficient in front of r:

$r = (3sin(theta))/(2cos^2(theta) -2cos(theta)sin(theta)-7sin^2(theta))$

Here is the graph the polar equation.

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

This proves that the conversion is done properly.

How do you convert 3y= 2x^2-2xy-7y^2 3y= 2x^2-2xy-7y^2 into a polar equation?