the curve h(t)=(t^3,t^2-t).show that it's a reg parametrized curve.?

Question

the curve g(t)=(-1+3t,2-3t) is a tangent of h(t). Find where h(t) and G(t) intersect. i.e. find the points such that g(t1)=h(t2) for some t1,t2, in the reals.

i've shown its a regular parametrized curve. i've tried finding the intersecting point by -1+3t=t^3 and 2-3t=t^2-t
but i'm not really gettin gud answers so im thinkn if im doin it wrong, not quite sure wat the question means by t1 and t2. ???

it also asks which intersection point is the tangent one? give reasons to ur answer.

Answer 1

t^3-3t+1=0 and
t^2+2t-2=0
they must have a common root which also must be a root of the remainder of its quotient
this remainder is 4t-4 =0 so t=1 which is not a root so h(t) and g(t) do not intercept
for h dx/dt= 3t^2 and dy/dt = 2t-1
for g dx/dt = 3 and dy /dt =-3 so the slope is -1
(2t-1)/3t^2=-1 so 3t^2+2t-1=0 and t = (-2+-sqrt(4+12))/6

t= -1 and t= 1/3
At these points (-1,2) and ( 1/27,-2/9) the tangent has slope-1
but for t=-1 the point on the tangent is (-4,1)
Thats only for sake of verification