Parametric equation of the parabola

[Blake Andrews]

Hi,

I have a question asking to write a set of parametric quations for the following parabola:

x²=-36y

I used:

x²=-4ay
x²=-4(9)y, so a=9

The parabola is in the form x²=-4ay so it is concave down with parametric co-ordinates of
x=2at and
y=-at²

So x=2(9)t
=18t
and
y=-9t²

My textbook is telling me the answer is x=-18t, y=-9t², which makes sense to me as it's concave down but i can't see where i went wrong in the method i used.

 

[HallsofIvy]

Well, if \(\displaystyle x= 18t\) then \(\displaystyle x^2= (18t)^2= 324t^2\) and if \(\displaystyle y= -9t^2\) so that \(\displaystyle x^2= -36y\) becomes \(\displaystyle 324t^2= -36(-9t^2)= 324t^2\). Yes, those are appropriate parametric equations.

So what about \(\displaystyle x= -18t\), \(\displaystyle y= -9t^2\)? Well, because x is squared, that sign doesn't matter! Do you understand that there can b infinitely many different parametric equations for the same curve?

 

[Steven G]

x²=-36y
Let y=-t.
So x² = 36t. Then x=6t^1/2.
As Prof Halls pointed out there are many solution. You did not have to introduce that number 4, but you could (and did) have.

 

[Blake Andrews]

Makes sense, thanks for your help.