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

Question

1466 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-26T01:25:04

$r = 0$ and
$r = ( 2 cos theta + 3 sin theta)/(2 cos^2 theta - sin^2 theta )$ .
This hyperbola passes through the pole $( O ), r = 0$ .

Conversion formula:

$( x. y ) = r ( cos theta, sin theta ), r = sqrt (x^2 + y^2) >= 0$

$2x^2 - y^2- 2x - 3y = 0$ converts to

$r = 0$ and

$r(2 cos^2 theta - sin^2 theta ) - ( 2 cos theta + 3 sin theta)) = 0$ .

This hyperbola passes through the pole $( O ), r = 0$ .

See graph.

graph{2x^2 - y^2- 2x - 3y = 0 [-20 20 -12 8] }

.

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