Question

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

Answer 1

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

Explanation:

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

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)`

Explanation:

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]}