The correct option is A a parabola
Given parameters are
x=t2+1⇒t2=x−1 ⋯(1)y=2t+1⇒t=y−12 ⋯(2)
From (1) and (2),
x−1=14(y−1)2⇒y2−2y−4x+5=0
Now comparing it with general form of conic, we get
a=0,b=1,h=0,c=5,g=−2,f=−1
Now, Δ=0+0−0−4−0=−4≠0
and h2=0=ab
Hence, given conic represents a parabola.