stupid general question regarding Parametrics (1 Viewer)

[ninetypercent]

Alright, I'm so confused.
For the parabola, x^2 = -4ay
one textbook said that the parametric equations are x = 2at, y = -at^2
however the jones and couchman textbook answers says that it is x = -2at, y = -at^2

both will give you the cartesian equation x^2 = -4ay, but only one works when u solve locus equations.

Which one is correct, btw?

thanks.

 

[Prashant Sallan]

I thought it was:

x=2at and y = at^2 :S

 

[jet]

I'm quite sure you only need to do any locus problems with the standard x² = 4ay.
But if I were going to choose a set I'd choose the first.

 

[ninetypercent]

thx guys

 

[Iruka]

Alright, I'm so confused.
For the parabola, x^2 = -4ay
one textbook said that the parametric equations are x = 2at, y = -at^2
however the jones and couchman textbook answers says that it is x = -2at, y = -at^2

both will give you the cartesian equation x^2 = -4ay, but only one works when u solve locus equations.

Which one is correct, btw?

thanks.

They are both correct as the parabola is symmetric about its axis.

But it is probably easier to use the first.

 

[ninetypercent]

ahh cool. say for example I'm deriving the equation of the tangent from the first.
that would be dy/dx = -x/2a
m = 2at/2a = t
t(x-2at) = y + at^2
tx - 2at^2 = y + at^2
tx - 3at^2 = y

and using the second one. dy/dx = -x/2a
m = -2at/2a = -t
-t (x+2at) = y+at^2
-tx - 2at^2 = y + at^2
-tx - 3at^2 = y

they give u two different results. and I think I tried using the second one in a locus question and it didn't work.

alright, I'll stick with the first one.

 

[Iruka]

What has happened is that the first representation has given you the tangent to a point on one side of the parabola, and the second has given you the tangent to a point on the other side of the parabola. The second equation is a mirror image of the first one. So the same point in cartesian coordinates on the parabola will correspond to different values of t, either positive or negative, depending on which representation of the curve you are using. You might want to try some examples using actual numbers for t and a, to see what I mean: Let a=1 and try to find the tangent at the point (2,-1). With the first representation, you use t=1. With the second, you have to use t=-1.


I think it is easier to see what is going on if you get points on the right hand side of the parabola when you use a positive value of t, and points on the left hand side when you use negative values of t. This is what the first parameterization gives you. In the second parameterization they are swapped around, which I think is a bit counter-intuitive, although not mathematically incorrect.