Parametric equations - intersection with parabola

[sambellamy]

I am asked to find the point at which the line defined by:
x = 1 + 3t
y = 2 - t

Intersects with the parabola:
y = x².

I am a bit confused on how to set up the equations to match. I have gathered that the y value has to be the same:
x² = 2 - t
x = (2 - t)^1/2

and then I plugged this into the x equation:
1 + 3t = (2 - t)^1/2
...
7t + 9t² = 1

What do I do next? I can factor out a t but I'll still have one inside.

Thanks in advance

 

[Steven G]

I am asked to find the point at which the line defined by:
x = 1 + 3t
y = 2 - t

Intersects with the parabola:
y = x².

I am a bit confused on how to set up the equations to match. I have gathered that the y value has to be the same:
x² = 2 - t
x = (2 - t)^1/2 Not true. For example if x^2 = 4 then x is not 4! If x^2 = 4 then x= +/- 2. In general, if x^2 = k, k>=0 then x = +/- sqrt(k)
and then I plugged this into the x equation:
1 + 3t = (2 - t)^1/2
...
7t + 9t² = 1

What do I do next? I can factor out a t but I'll still have one inside.

Thanks in advance

Why not eliminate t from the 1st two eqs?
x = 1 + 3t, then (x - 1)/3 =t
So y = 2-(x-1)/3
Simplify the y equation in the last line, set equal to x^2 and determine the point of intersection.

Show us how you make out.

 

[stapel]

I am asked to find the point at which the line defined by:
x = 1 + 3t
y = 2 - t

Intersects with the parabola:
y = x².

I am a bit confused on how to set up the equations to match. I have gathered that the y value has to be the same:
x² = 2 - t

I think the general idea is to get rid of the parameter "t". So try solving for that, and then substituting to eliminate:

. . . . .x = 1 + 3t

. . . . .x - 1 = 3t

. . . . .(x - 1) / 3 = t

Then:

. . . . .y = 2 - t = 2 - [ (x - 1) / 3 ]

. . . . .y = 2 - x/3 + 1/3

. . . . .y = (-1/3)x + 7/3

So you've got a straight line. Now set this equal to the parabola:

. . . . .(-1/3)x + 7/3 = x²

...and solve:

. . . . .0 = x² + (1/3)x - 7/3

Plug into the Quadratic Formula, and so forth.

 

[Ishuda]

...
7t + 9t² = 1

What do I do next? I can factor out a t but I'll still have one inside.

Thanks in advance

You can do it this way with just a few more steps. First, you have a quadratic equation in t, i.e.
9 t² + 7 t - 1 = 0.
The solutions for t are found by using the quadratic equation and to find x and y plug those answers back into the original equations.

 

[Steven G]

I am asked to find the point at which the line defined by:
x = 1 + 3t
y = 2 - t

Intersects with the parabola:
y = x².

I am a bit confused on how to set up the equations to match. I have gathered that the y value has to be the same:
x² = 2 - t
x = (2 - t)^1/2 not true!. If x^2 = k, k>0, then x = +/- sqrt(k). So x=+/- (2 - t)^1/2 . In the end you were ok since you squared (2 - t)1/2. A mistake that cancels out is still a mistake
and then I plugged this into the x equation:
1 + 3t = (2 - t)^1/2
...
7t + 9t² = 1

What do I do next? I can factor out a t but I'll still have one inside.

Thanks in advance

.