Parametric Equation problems?

Question

Show that the curve x = t^2 - 3t + 5, y = t^3 +t^2 -10t + 9 intersects itself at the point (3,1), and find equation for two tangent lines to the curve at the point of intersection.

Answer 1

Set x=3, then solve x's equation for t.

x: t^2 - 3t + 5 = 3 becomes t^2 - 3t + 2 = 0.
Factoring gives (t-1)(t-2) = 0, so t=1 or t=2.

Now plug t=1 and t=2 into the equation for y. In both cases, you get y=1, so both t=1 and t=2 correspond to the point (3,1).

Now, compute dy/dx = (dy/dt) / (dx/dt). You get

dy/dx = (3t^2 + 2t - 10) / (2t - 3).

Plug in t=1 and t=2 to get the slopes of the two tangent lines. For each slope m, use point-slope form

y - 1 = m(x - 3)

to get the equations of the tangent lines.

Answer 2

all you need to do
is to see that
x= t^2 - 3t + 5 =3 has 2 solutions
and that
y = t^3 +t^2 -10t + 9 = 1 has 2 solutions,
the two solutions of x and the 2 of y are the same.
in both cases you get t=1 and t=2

Now, compute dy/dx = y' / x'. You get
dy/dx = (3t^2 + 2t - 10) / (2t - 3).
substitute t=1 and t=2 to get the slopes of the two tangent lines.
m1=3+2-10 / 2 -3 = -5/-1=5
m2=12+4-10 / 4-3 = 6/1= 6

y - 1 = 5(x - 3)
and the other one:
y - 1 = 6(x - 3)