The correct option is B Parabola
x=t2+t and y=2t−1
By elimination of t
t=y+12
x=(y+12)2+y+12
y2+4y−4x+3=0
On comparing with ax2+by2+2hxy+2gx+2fy+c=0
we have a=0,b=1,g=−2,f=2,h=0,c=3
Δ=abc+2fgh−af2−bg2−ch2=−4≠0
and h2−ab=0⇒h2=ab
So, the given equation is a parabola