00:01
Okay, so in order to find the intersection points, we want to find t1 and t2, which are not equal to each other, such that x evaluated at t1 is equal to x evaluated at t2 and the same for the y's, because then these two values of t2 both give the same coordinates.
00:27
So that's why we get an intersection.
00:32
To 1 minus t1 squared equals 1 minus t2 squared and t1 minus t1 cubed equals t2 minus t2 cubed.
00:44
Now from the top equation, we can see that the ones cancel and we find that t2 is just equal to plus or minus t1.
00:55
And if we plug this into here, we're going to find that we get t1 minus t1 cubed is equal to, now if we plug in the plus, version, it's just going to give us the exact same thing, so that's not going to give us any condition.
01:10
So if we plug in the minus version, then we're going to get minus t1 plus t1 cubed.
01:19
And we can rearrange this to find that 2t1 minus 2t1 cubed or times 1 minus t1 squared is equal to 0.
01:30
And so t1 has to equal 0 or plus or minus 1.
01:35
And using the fact that this was when t2 is equal to minus t1, this implies that t2 is equal to zero or minus or plus one.
01:50
However, we can see that for this solution, the t's are not distinct, so this doesn't give us a point of intersection, but for these ones, it does...