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

Question

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

Brooklyn Anderson · Accepted Answer

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

for polar conversion

#r^2=x^2+y^2#

#x=rcostheta#

#y=rsintheta#

for

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

substituting from above.

#rsintheta=rcostheta-2rsintheta+r^2cos^2theta-3r^2sin^2theta#

cancelling and rearranging for #""r#

#cancel(r)sintheta=cancel(r)costheta- 2cancel(r)sintheta+cancel(r^2)^rcos^2theta-3cancel(r^2)^rsin^2theta#

#sintheta=costheta-2sintheta+rcos^2theta-3rsin^2theta#

#sintheta-costheta+2sintheta=rcos^2theta-3rsin^2theta#

#sintheta-costheta+2sintheta=r(cos^2theta-3sin^2theta)#

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

Oliver Atkinson · Answer

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

The conversion formula is #(x, y)=r(cos theta, sin theta)#

Making substitutions and reorganizing for the explicit form

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

This represents a hyperbola, with center at (-1/2, -1/2). See graph.

graph{x^2-3y^2+x-3y=0 [-2.5, 2.5, -1.25, 1.25]}

Abigail Armstrong · Answer

undefined To convert the equation $ y = x - 2y + x^2 - 3y^2 $ into a polar equation, you'll need to express $ x $ and $ y $ in terms of $ r $ and $ \theta $, where $ r $ represents the distance from the origin and $ \theta $ represents the angle from the positive x-axis.

Given that $ x = r \cos(\theta) $ and $ y = r \sin(\theta) $, substitute these expressions into the given Cartesian equation:

$ y = x - 2y + x^2 - 3y^2 $

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

Now, simplify and solve for $ r $ in terms of $ \theta $ to obtain the polar equation.