Good evening. How would I find the parametric equation of a parabola given three points? My point are endpoints $(-1, 2)$ and $(3, 4)$. However, the curve must also pass through $(0,0)$.
I have tried to solve in various ways using $x = at^2$ and $y = 2at$ but I keep going around in circles when I solve. I am really stumped.
Thank you so much in advance for your help.
Gabrielle
On
There are an infinite number of parabolas passing through these three points, so I’ll proceed assuming that the parabola’s axis is meant to be parallel to the $y$-axis.
Plug your points into the general equation $y=ax^2+bx+c$ one by one and solve for the unknown coefficients. So, you have $$a(-1)^2+b(-1)+c=a-b+c=2,$$ which means $c=2-a+b$. Next, you have $$a(0)^2+b(0)+c=2-a+b=0,$$ from which $b=a-2$ and thus $c=0$. Finally, $$a(3)^2+b(3)+c=9a+3(a-2)+0=12a-6=4$$ which yields $a=5/6$.
I expect that you’ll be able to come up with some parameterization from this Cartesian equation on your own.
There are infinte parabolas passing through three points. But none of these have the $y$-axis as an axis.
The general equation of a parabola is $(Ax + By)^2+Cx+Dy+E=0$. Substituting $(-1,2)$, $(0,0)$, $(3,4)$, you can get any of the infinite number of solutions.