Question

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

Answer 1

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

Explanation:

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

Answer 2

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

Explanation:

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..