How do you convert #8=(3x-y)^2+y-5x# into polar form?

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Answer 1 · 2016-12-09T16:26:35

#r^2(3cos theta-sin theta)^2+r(sin theta-5cos theta)-8=0# representing a small-size parabola-

As the second degree terms form a perfect square, the equation

represents a parabola.

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

Substitutions give the form given in the answer,

It is incredible but true. This equation would reduce to the simple

form

#2a/r=1+cos theta#

referred to the focus as pole ( r = 0 ) and the axis, #rarr # vertex, as

#theta = 0#. Here, ( the focus-vertex distance ) a is the size of the

parabola

graph{(3x-y)^2+y-5x-8=0 [-40, 40, -20, 20]}