y=√1+2x, x≥0
y=√1+4x, x≥0
x=√1+4y, y≥0
x=√1+2y, y≥0
The correct option is y=√1+2x, x≥0√h²+k²=|h|+1 ⇒x²+y²=x²+1+2x ⇒y²=1+2x ⇒y=√1+2x; x≥0