Points of intersection of Parametric Lines

Dramen
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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
 
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Dramen said:

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.
 
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)?
 
Dramen said:
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.
 
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
 
Dramen said:
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.
 
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)
 
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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?
 
Yep I checked it and the numbers work.
Thanks for the nudges in the right direction.
 
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