How do you convert 3y= 2x^2-2xy-7y^2 3y=2x2−2xy−7y2 into a polar equation?

1 Answer
Douglas K.
Mar 30, 2017

Substitute rcos(theta)rcos(θ) for xx and rsin(theta)rsin(θ) for y.
Write rr as a function of thetaθ

Explanation:

Given: 3y= 2x^2-2xy-7y^23y=2x2−2xy−7y2

Here is the graph of the Cartesian equation:

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

Substitute rcos(theta)rcos(θ) for xx and rsin(theta)rsin(θ) for y.

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

Write rr as a function of thetaθ

3sin(theta)r = (2cos^2(theta) -2cos(theta)sin(theta)-7sin^2(theta))r^23sin(θ)r=(2cos2(θ)−2cos(θ)sin(θ)−7sin2(θ))r2

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

3sin(theta) = (2cos^2(theta) -2cos(theta)sin(theta)-7sin^2(theta))r3sin(θ)=(2cos2(θ)−2cos(θ)sin(θ)−7sin2(θ))r

Divide by the coefficient in front of r:

r = (3sin(theta))/(2cos^2(theta) -2cos(theta)sin(theta)-7sin^2(theta))r=3sin(θ)2cos2(θ)−2cos(θ)sin(θ)−7sin2(θ)

Here is the graph the polar equation.

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

This proves that the conversion is done properly.

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