How do you convert #9=(x-4y)^2-2y+x# into polar form?

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Answer 1 · 2018-08-12T20:02:00

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

#x = r cos theta, y = r sin theta#

Answer 2 · 2018-08-13T00:44:31

The conversion has been done in the answer by the questioner.
This is added for the information that the graph is a parabola.

The second degree terms form a perfect square, and so, this

represents a parabola. See graph'
graph{((x-4y)^2-2y+x-9)(y-1/4x-0.071)(y+4x-77.1)=0[17 19 4 5 ] }

Referring to its focus #S# as pole r = 0 and the

perpendicular from S on the directrix in the direction as the initial

line (#theta = 14.036^o#), the polar equation reads

#(2a)/r = 1 + cos (theta-14.036^o )
.
From the given equation, the slope of the axis is 1/4. The angle =

#14.036^o#.

The axis: y = 1/4x=0.071.

The tangent at the vertex is y = - 4x + 77.1, and the vertex is (

18.125, 4.6 ), nearly.

Once again, I emphasize that this graphic utility has high potential,

for the Cartesian frame. My data are graphic approximations. The

precision could be higher..