The correct option is A x=y2−y+2
vx=8t−2 or, dxdt=8t−2
Integrate to get, x=4t2−2t+c
Given, at t=2, x=14, thus 14=4(4)−2(2)+c or c=2
Thus, x=4t2−2t+2
Similarly, vy=2 or dydt=2
Integrate to get, y=2t+c′
Given, at t=2, y=4, thus 2(2)+c′=4 or c′=0
Thus, y=2t (or t=y2)
Put this t in x to get x=4(y2)2−2(y2)+2=y2−y+2