If I have three collinear points in the $x$-$y$ plane, $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, and each has an associated point in the $z$ direction, how do I derive the equation for a parabola that passes through each of the $z$ points?

Also, is there any way to prove that if the parabola was projected on to the $x$-$z$ or $y$-$z$ planes, that the resulting curve would also be a parabola (I have a feeling it would be)?


  • $\begingroup$ How is it a $1$-D parabola? Isn't it a $2$-D parabola? $\endgroup$ Dec 30, 2017 at 17:38
  • $\begingroup$ Good point, I'll try to edit $\endgroup$
    – D. W.
    Dec 30, 2017 at 17:39
  • $\begingroup$ If you move about your axes about you will find that it is simply a normal parabola with $z = 0$. $\endgroup$ Dec 30, 2017 at 17:41
  • $\begingroup$ And yes it should also be a parabola as all you are doing is shifting the axes. $\endgroup$ Dec 30, 2017 at 17:44
  • $\begingroup$ $z=0$? Then I would just have the three points in the $x-y$ plane they are collinear... Do you mean rotating about $z$-axis? $\endgroup$
    – D. W.
    Dec 30, 2017 at 17:47

1 Answer 1


It is not clear from the question what kind of equation is expected. If $y=mx+q$ is the equation of the line in the $x-y$ plane, I think the most natural way to handle such a parabola is that of finding the equation of its projection on the $x-z$ plane, which is a parabola of equation $$ \tag{1} z=ax^2+bx+c. $$ You can find $a$, $b$ and $c$, as usual, by plugging there the coordinates of the three given points: only $x$ and $z$ are needed, because $y=mx+q$.

In other words: the parabola is the intersection between the parabolic cylinder given by equation $(1)$ and the plane of equation $y=mx+q$.

If the line in the $x-y$ plane is parallel to the $y$ axis then you can instead project the parabola onto the $y-z$ plane.

  • $\begingroup$ The equation of the curve through 3-space is what I was looking for for the first part of the question $\endgroup$
    – D. W.
    Dec 30, 2017 at 18:15
  • $\begingroup$ How do we know that the projection on to the x-z plane will be a parabola? $\endgroup$
    – D. W.
    Dec 30, 2017 at 18:18
  • $\begingroup$ Watch this youtube.com/watch?v=hoh4TmPzu1w $\endgroup$ Dec 30, 2017 at 18:37
  • $\begingroup$ @D.W. We know because you can imagine your parabola as the intersection of the parabolic cylinder $z=ax^2+bx+c$ with plane $y=mx+q$. The intersection of a quadric with a plane is a conic, which in the case at hand is not limited and is made of a single piece, hence a parabola. $\endgroup$ Dec 30, 2017 at 19:08

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