Find parametric equations xxt yyt for a particle moving alo

Find parametric equations x=x(t) ; y=y(t) for a particle moving along the parabolic path y=-2(x-2)^2 if the particle is at the vertex at t=3 sec and at the point (4,-8) at t=7 sec. Please help by posting all the work!

Solution

Solution: y = -2(x-2)^2 ,

y = - 4(1/2)(x-2)^2, therefore a = 1/2

We know that the perametric equation of a parabola is

x = h - at^2, y = 2at.

Thus

The vertex of the parabola is (2,0)

At t = 3, x = 2,

    2 = h - (1/2)(3)^2

or   2 + (9/2) = h or h = 13/2,

At t = 7, x = 4, y = - 8,

Thus y = 2 (1/2)t = t

or      (4-2)^2 = 4

Therefore the parametric equations x = 13/2+(1/2)t,   y = t