3
$\begingroup$

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)?

1D3DParab

$\endgroup$
  • $\begingroup$ How is it a $1$-D parabola? Isn't it a $2$-D parabola? $\endgroup$ – Mohammad Zuhair Khan Dec 30 '17 at 17:38
  • $\begingroup$ Good point, I'll try to edit $\endgroup$ – D. W. Dec 30 '17 at 17:39
  • $\begingroup$ If you move about your axes about you will find that it is simply a normal parabola with $z = 0$. $\endgroup$ – Mohammad Zuhair Khan Dec 30 '17 at 17:41
  • $\begingroup$ And yes it should also be a parabola as all you are doing is shifting the axes. $\endgroup$ – Mohammad Zuhair Khan Dec 30 '17 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 '17 at 17:47
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '17 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 '17 at 18:18
  • $\begingroup$ Watch this youtube.com/watch?v=hoh4TmPzu1w $\endgroup$ – Mohammad Zuhair Khan Dec 30 '17 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. $\endgroup$ – Aretino Dec 30 '17 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.