eliminate parameter to find cartesian equation!?

Given:

x(t) = t^2 + t

y(t) = t^2 - t

Solve your x-equation for t, via quadratic formula:

t = -1/2±1/2*sqrt(1+4*x)

Substitute both versions in to y:

y = ( -1/2±1/2*sqrt(1+4*x))^2 - ( -1/2±1/2*sqrt(1+4*x))

Simplify:

y = 1 ± sqrt(1+4*x) + x

In order to remove the pesky ± sign, re-arrange with the anticipation of squaring that term:

y - x - 1 = ± sqrt(1+4*x)

(y - x - 1)^2 = 1 + 4*x

Expand and gather to one side:

y^2 - 2*y*x - 2*y + x^2 - 2*x = 0

So you can either conclude:

y^2 - 2*y*x - 2*y + x^2 - 2*x = 0

or

y = 1 ± sqrt(1+4*x) + x

This is the equation of a rotated parabola.

x=t^2+t ---------------(1)

y=t^2-t ----------------(2)

Now from(1)+(2)

x+y=2.t^2 ----------(3)

from (1)-(2)

x-y=2.t -----------(4)

or t=(x-y)/2 substitute this in (3)

(x+y)=2.(x-y)^2/4

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

If you want can simplify further otherwise leave as it is.

IVAN