# How do you graph an equation in x and y that has the same graph as the polar equation r = 4sin(theta) - 2cos(theta)I need to be able to find the equation in x and y before graphing.

embizze | High School Teacher | (Level 1) Educator Emeritus

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Given `r=4sintheta-2costheta`

We have the following conversions:

`x=rcostheta==>costheta=x/r`

`y=rsintheta==>sintheta=y/r`

`r^2=x^2+y^2`

Then:

`r=4sintheta-2costheta` substitute for sin and cos

`r=(4y)/r-(2x)/r`  Multiply both sides by "r"

`r^2=4y-2x` substitute for `r^2`

`x^2+y^2=4y-2x` Write in standard form:

`x^2+2x+y^2-4y=0` Use complete the square:

`(x^2+2x+1)+(y^2-4y+4)-1-4=0`

`(x+1)^2+(y-2)^2=5`

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The equation in polar form `r=4sintheta-2costheta` is equivalent to the equation in cartesian form `(x+1)^2+(y-2)^2=5` which is a circle centered at (-1,2) with radius `sqrt(5)` .

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The graph of `r=4sintheta-2costheta` :

The graph of `(x+1)^2+(y-2)^2=5` :

(This is a circle -- the grapher only graphs functions and a circle does not represent a function in cartesian coordinates. Thus you have to break the graph into two functions: `y=2+-sqrt(5-(x+1)^2)` thus causing the "gaps")

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