How do you convert #3= -x^2+3xy-4y^2 # into a polar equation?

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Answer 1 · 2016-12-21T12:34:53

#r = 1/3sqrt((cos theta-sin theta)(2 sin theta - cos theta))#.

#-3=(x-2y)(x-y)# represents a hyperbola having asymptotes

x = 2y and x = y#

Use conversion formula # (x, y)= r (cos theta, sin theta )#.

The polar form is

#3= ((-cos^2theta+3sin theta cos theta -4 sin^2theta)/r^2#

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

Explicitly,

#r = 1/3sqrt((cos theta-sin theta)(2 sin theta - cos theta))#

The asymptotes are now obtained using r = 0 at the ( meet of the

asymptotes ) center..

So, they are ( for pairs of opposite directions ) #theta = pi/4, 5/4pi #

and #theta = pi+tan^(-1)(1/2), pi + tan^(-1)(1/2)#

Note that #theta# for the hyperbola #in (tan^(-1)(1/2), pi/4)# and

#(pi+tan^(-1)(1/2), 5/4pi)#, for the respective branches, in #Q_1 and Q_3#.

graph{(x-y)(x-2y)+3=0 [-40, 40, -20, 20]}