Question

How do you convert the parametric equations into a Cartesian equation by eliminating the parameter r: `x=(r^2)+r`, `y=(r^2)-r`?

Answer 1

`x^2+y^2 -2x-2y -2xy = 0`

Explanation:

We have:

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

Adding the equations:

`x+ y = 2r^2 => r^2 = 1/2(x+y)`

Multiplying the Equations we get:

`xy = (r^2 + r)(r^2 - r)`
`\ \ \ \ = r^4 - r^2`

And substituting `r^2 = 1/2(x+y)` gives:

`xy = (1/2(x+y))^2 - 1/2(x+y)`

Thus the Cartesian equation is:

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

Answer 2

`x^2+y^2-2xy-2x-2y = 0`

Explanation:

We have:

`{(x=r^2+r),(y=r^2-r):}`

Summing the equations we have:

`x+y = 2r^2`

and subtracting the second from the first:

`x-y = 2r`

or:

`r=(x-y)/2`

Then:

`x+y = 2((x-y)/2)^2`

`x+y = (x-y)^2/2`

`2x+2y = x^2-2xy+y^2`

and finally:

`x^2+y^2-2xy-2x-2y = 0`

This is the equation of a conic, so we can calculate the invariants:

`det ((1,-1,-1),(-1,1,-1),(-1,-1,0)) = -4`

The cubic invariant is non null so the conic is non-degenerate,

`det( ( 1,-1),(-1,1)) = 0`

the quadratic invariant is null so the conic is a parabola.

graph{x^2+y^2-2xy-2x-2y=0 [-213.9, 213.8, -106.2, 107.5]}