Question

How do you convert `3y= 3x^2-6xy-x` into a polar equation?

Answer 1

`r=sec theta( (3tan theta+1)/(3-6tan theta))`. The graph of the given hyperbola is inserted.

Explanation:

The conversion formula is

`(x, y)=r(cos theta, sin theta)`,

Rearranging, the given equation becomes

`x= r cos theta)=(3y-x)?(3x-6y)`

`=(3 sin theta - cos theta)/(3cos theta-6 sin theta)`

`=(3 tan theta+1)(3-6 tan theta)`, So, explicitly,

`r=sec theta( (3tan theta+1)/(3-6tan theta))`

graph{3x^2-6xy-3y-x=0 [-10, 10, -5, 5]}