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

Question

1555 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-05T05:06:37

$r = sec θ (\frac{3 tan θ + 1}{3 - 6 tan θ})$ $r=sec theta( (3tan theta+1)/(3-6tan theta))$ . The graph of the given hyperbola is inserted.

The conversion formula is

$(x, y) = r (cos θ, sin θ)$ $(x, y)=r(cos theta, sin theta)$ ,

Rearranging, the given equation becomes

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

$= \frac{3 sin θ - cos θ}{3 cos θ - 6 sin θ}$ $=(3 sin theta - cos theta)/(3cos theta-6 sin theta)$

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

$r = sec θ (\frac{3 tan θ + 1}{3 - 6 tan θ})$ $r=sec theta( (3tan theta+1)/(3-6tan theta))$

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

How do you convert 3y= 3x^2-6xy-x 3y=3x2−6xy−x3y= 3x^2-6xy-x into a polar equation?