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

Question

1302 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-03T11:48:11

Polar equation of hyperbola is ,
#r =(3 cos theta - sin theta)/(sin^2 theta-3 cos ^2 theta) #

In rectangular - polar conversion,

#r^2=x^2+y^2 , x= r cos theta , y= r sin theta #

#y= 3 x^2+3 x-y^2# or

#y^2 + y= 3 x^2+3 x# or

#y(y+1)= 3 x(x+1)# or

# cancelr sin theta (r sin theta +1)= 3 cancelr cos theta( r cos theta+1)# or

#r sin^2 theta+sin theta = 3 r cos ^2 theta+3 cos theta# or

#r sin^2 theta-3 r cos ^2 theta=3 cos theta - sin theta# or

#r (sin^2 theta-3 cos ^2 theta) =3 cos theta - sin theta# or

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

Polar equation of hyperbola is ,

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

graph{y=3 x^2+3 x-y^2 [-10, 10, -5, 5]} [Ans]