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

2 Answers
May 18, 2017

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

May 18, 2017

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]}