What is the cartesian equation for the polar curve # r = 2sin theta - 4cos theta #?

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Answer 1 · 2017-07-07T15:55:09

#x^2 + y^2 = 2y +- 4sqrt(x^2+y^2)#

Use the conversions:

#r^2 = x^2 + y^2#

#r = +-sqrt(x^2+y^2)#

#rsintheta = y#

#rcostheta = x#

First, let's multiply both sides of the equation by #r#.

#r*r = r*(2sintheta-4)#

#r^2 = 2rsintheta - 4r#

Now we can substitute the rectangular forms.

#x^2 + y^2 = 2y - 4(+-sqrt(x^2+y^2))#

#x^2 + y^2 = 2y +- 4sqrt(x^2+y^2)#

We can simplify this further, but this is a good stopping point.

Final Answer

Answer 2 · 2017-07-10T14:39:17

The cartesian rectangular equation is:

# x^2 + 4x + y^2 - 2y = 0 #

Which can also be written as:

# (x+2)^2 + (y-1)^2 = sqrt(5)^2#

Which is a circle centred on #(-2,1)# with radius #sqrt(5)#

Assuming the correct equation to be:

# r = 2sin theta - 4cos theta #

To convert to polar coordinates to cartesian rectangular form, using the relationships:

# {: (x = rcos theta, cos theta=x/r), (y = rsin theta, sin theta=x/r) :} } => r^2 = x^2+y^2 #

So we can write the polar equation as follows:

# r = 2sin theta - 4cos theta #
# \ = 2(y/r) - 4(x/r) #

Multiply by #r# and we get:

# r^2 = 2y - 4x #

# :. x^2+y^2 = 2y - 4x #

# :. x^2 + 4x + y^2 - 2y = 0 #

Although this would be sufficient, we can analyze a little further by completing the square on both #x# and #y#:

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

# :. (x+2)^2 -4 + (y-1)^2 - 1 = 0#
# :. (x+2)^2 + (y-1)^2 = 5#

# :. (x+2)^2 + (y-1)^2 = sqrt(5)^2#

Which is a circle centred on #(-2,1)# with radius #sqrt(5)#

graph{x^2 + 4x + y^2 - 2y = 0 [-10, 10, -5, 5]}