\(\displaystyle x = 3cos^2(t)+3, \ y \ = \ 2sin^2(t)-1\)
\(\displaystyle cos^2(t) \ = \ \frac{x-3}{3}, \ sin^2(t) \ = \ \frac{y+1}{2}\)
\(\displaystyle sin^2(t)+cos^2(t) \ = \ \frac{y+1}{2} \ + \ \frac{x-3}{3}\)
\(\displaystyle Ergo, \ \frac{y+1}{2} + \ \frac{x-3}{3} \ = \ 1\)
