Points of intersection of Parametric Lines

[Dramen]

Homework Statement

I'm told to find the 2 points the two curves P and Q will intersect on and the parametric equations are:
P (x=t, y=2t-1)
Q (x=3t-t^2, y=t+1)

The Attempt at a Solution

I know I'm supposed to set x-equations and y-equations equal to each and solve so that

t=3t-t^2 for x
2t-1=t+1 for y

and when I solve them I get t=0 and t=2 for x
and
t=2 for y
the problem is I can't seem to find another t-value for y

Also I'm not completely sure if I can use t interchangeably between the two equations when solving for them or if I should consider the t to be two separate and unique variables like t_1 and t_2

 

[LCKurtz]

Homework Statement

I'm told to find the 2 points the two curves P and Q will intersect on and the parametric equations are:
P (x=t, y=2t-1)
Q (x=3t-t^2, y=t+1)

The Attempt at a Solution

I know I'm supposed to set x-equations and y-equations equal to each and solve so that

t=3t-t^2 for x
2t-1=t+1 for y

and when I solve them I get t=0 and t=2 for x
and
t=2 for y
the problem is I can't seem to find another t-value for y

Also I'm not completely sure if I can use t interchangeably between the two equations when solving for them or if I should consider the t to be two separate and unique variables like t_1 and t_2

That is the problem. You can't assume the curves cross for the same value of the parameters. So call one parameter $t$ and the other $s$ and try setting the $x$ values and $y$ values equal.

 

[Dramen]

I did just that quite a while ago when my instructor had hinted at that idea and this is what I came up with

t=3s-s^2
t=s(3-s)
so that for x t=s and t=3-s
so that any same two values fit in the first equality(?) and only 1.5 solves the equality in the second

for y it is
2t-1=s+1
2t=s+2
so that for y t=2 and s=2

I'm still not sure how to find a second point of intersection. Since the first point is (1.5,2)?

 

[LCKurtz]

I did just that quite a while ago when my instructor had hinted at that idea and this is what I came up with

\color{red}{t=3s-s^2}
t=s(3-s)
so that for x t=s and t=3-s
so that any same two values fit in the first equality(?) and only 1.5 solves the equality in the second

for y it is
\color{red}{2t-1=s+1}
2t=s+2
so that for y t=2 and s=2

I'm still not sure how to find a second point of intersection. Since the first point is (1.5,2)?

Ignoring the other bad logic, you have two equations in two unknowns. Solve them correctly.

 

[Dramen]

I'm no good when trying to solve for 2 unknowns algebraically like this, because first thought is to substitute t=3s-s^2 into 2t-1=s+1 but that won't work. And by graph for the first equation t=s=0 or 2 and the second t=s=2

 

[LCKurtz]

I'm no good when trying to solve for 2 unknowns algebraically like this, because first thought is to substitute t=3s-s^2 into 2t-1=s+1 but that won't work.

Yes it will. Try it.

 

[Dramen]

ok I did that so that 2(3s-s^2)-1=s+1
setting it to 0 gives me -2s^2+5s-2=0
and solving for that I get s=1/2 and s=2

then I plug those answers into t=3s-s^2
so that t=1.25 and t=2

so then my points of intersection are at
P (1.25,1.5) and (2,3)
Q (1.25,1.5) and (2,3)

 

[LCKurtz]

So check your work. Do your s and t values work in their equations? If so, do the two curves go through those two points for the corresponding values of s and t?

 

[Dramen]

Yep I checked it and the numbers work.
Thanks for the nudges in the right direction.