Equation of parabola in x-y plane

Question

Given the following equations:

$$x=t^2$$

$$y=t^2$$

$$z=2t+3$$

Show that the graph of the curve $(t^2,t^2,2t+3)$ lives in a plane and that, within that plane, the graph is a parabola.

My attempt, so far:

From the first two equations $x=y$, meaning the curve belongs to the $x-y=0$ plane.

Plugging the first or second equation into the third one:

$$((z-3)/2)^2=x,\>\>\> ((z-3)/2)^2=y$$

This tells me it is parabola in the zx and zy planes respectively , I guess ??

A matlab plot shows it is , in fact, a parabola in the $x-y=0$ plane. What I am doing wrong and how do I show it?

Answer 1

Rotate the $xyz$-coordinates by 45 degrees around the $z$ axis to the $uvz$-coordinates, which is equivalent to

$$x=u+v, \>\>\> y=u-v$$

Rewrite the curve in the $uv$-coordinates

$$u+v=t^2, \>\>\>u-v=t^2$$

Then, we have $u=t^2$ and $v=0$. So, the curve lies in the plane $v=0$, with the $uz$ coordinates given by,

$$u = t^2$$ $$z=2t+3$$

Eliminate $t$ to get the standard parabola equation in the plane $v=0$ (or, the plane $x=y$ in the original coordinates)

$$(z-3)^2=4u$$

Answer 2

If a curve $u(t)$ lies on a plane, then the plane defined by any three points $A$, $B$, $C$ on the curve must be always the same, i.e. the vector product $\vec n=(B-A)\times(C-A)$ must have the same direction for any choice of the points.

In practice we can keep two points fixed and choose, for instance: $A=u(0)=(0,0,3)$, $B=u(1)=(1,1,5)$, $C=u(t)=(t^2,t^2,2t+3)$, to find: $$ \vec n=(1,1,2)\times(t^2,t^2,2t)=(2t-t^2,2t-t^2,0)=(2t-t^2)(1,1,0). $$ The direction $\vec n/|\vec n|$ is then ${1\over\sqrt2}(1,1,0)$ and does not depend on $t$: hence the curve lies on a plane whose normal unit vector is ${1\over\sqrt2}(1,1,0)$.