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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

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    What are “endpoints” of a parabola? – 2017-01-15
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    I think my teacher means the domain – 2017-01-15

2 Answers 2

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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.

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    Can you please elaborate on how to perform the substitution - I am very confused about what to substitute in plus shouldn't myequations be in the form X= Y=. – 2017-01-14
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    Substituting $x=0$, $y=0$ gives $E=0$. $x=-1$, $y=2$ gives $(-A+2B)^2-C+2D=0$. $x=3$, $y=4$ gives $(3A+4B)^2+3C+4D=0$. This system of equations for $A$, $B$, $C$, $D$ has infinite solutions. – 2017-01-14
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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.