What the is the polar form of #y = (y-x)/(x+y) #?

Question

1202 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-06T02:44:51

#r = tan ( theta - pi/4 ) csc theta#

Cross multiplication gives a second degree equation

#( x + y - 2 )( y + 1 ) = -2#

that represents a hyperbola, with asymptotes

#( x + y - 2 )( y + 1 ) = 0#.

Using #( x, y ) = r ( cos theta, sin theta )#,

#r sin theta = (r(sin theta - cos theta ))/(r ( sin theta + cos theta ))#

giving

#r sin theta = ( tan theta - 1 ) / ( tan theta + 1 )#

#= ( tan theta - tan (pi/4 ))/( 1 + tan theta tan (pi/4) )#

#= tan ( theta - pi/4 )#

See graph for the hyperbola and asymptotes

#r = -csc theta#

and #r = 2/( cos theta + sin theta ) = sqrt2csc ( theta + pi/4)#
graph{ (y - (y-x)/(y+x))(y+1)(x+y-2)=0[-5 15 -6 4]}