Elipses Kreives simetrijos centras, uzdavinys

Sveiki turiu problema isspresti viena uzdavini, niekaip nesigauna, gal galite paaiskinti kaip ji isspresti?

https://scontent-ams3-1.xx.fbcdn.net/hphotos-xpa1/v/t34.0-12/12277131_466993843481026_1338160984_n.jpg?oh=5b265180e10457012332ced4d22581df&oe=564E6F63

x²+3x  +2y²+4y=-1
(x²+3x+9/4) -9/4    +2(y²+2y+1) -2=-1
(x+3/2)²  + 2(y+1)²=13/4
(x+3/2)²/(13/4)  + (y+1)²/(13/8)=1

vadinasi centras is (0;0) pasitraukia i (-1,5;-1)=(x0;y0)
vadinasi -1,5+-1=-2,5. arba -2,500