math 30
posted by Sandy on .
I need to write 4x^2  y^2 8x 4y + 16 = 0 in standard form and I keep getting it wrong. Could someone please help me. Thanks

4(x^22x+1) (y^2 +4y+4)=16
4(x1)^2(y+2)^2=(4)^2
(x1)^2/1 (y+2)^2/2^2=1
Now this negative sign presents a problem, so
(y+2)^2/4^2 (x1)^2/2=1
check that.