# The homework had first to do with the intersection of... paraboloid z = x + y

```The homework had first to do with the intersection of two surfaces, one, the
paraboloid z = x2 + y 2 , the other, the elliptic paraboloid z = 2x2 + 3y 2 − 1.
Consider the intersection of these surfaces. The second one dips below the first
when x and y are small, because of the −1 in its formula, but for large x and
y, the second surface is higher.
This means that the intersection is some sort of bounded thing. If we go
now to algebra, the intersection is the set of all (x, y, z) so that both equations
hold, at the same time, with the same values of the variables. Now if they both
hold, with the same z in each, then the things that are equal to z are equal to
each other. So x2 + y 2 = 2x2 + 3y 2 − 1 for any point (x, y, z) in the intersection.
That means that x2 + 2y 2 = 1 for any (x, y, z) in the intersection. And of
course, we must still have z = x2 + y 2 , for if both original equations must hold,
then this one in particular must hold.
It’s not that hard to find some
√ points that belong to both of the second pair
of curves: (&plusmn;1, 0, 1) and (0, &plusmn;1/ 2, 1/2) all work. If the curve lies in some plane,
then intuitively it must lie in a plane parallel to the x-y plane, so you’d answer
no it isn’t an ellipse, because these points aren’t all level with each other. But
if you want proof, it’s right to hand: these four points are not coplanar because
the line joining the first pair is skew to the line joining the second pair!
What about the second question of this problem: is the projection of the
intersection onto the x-y plane an ellipse? Projecting onto that plane means
replacing every point (x, y, z) in the intersection with the corresponding (x, y, 0).
But now, those points are just the ones satisfying x2 + 2y 2 = 1, z = 0. That’s
a classic ellipse in the x-y plane. So yes, this time it really is an ellipse.
The other curve? This shows them both.
1
This brings us to the other problem. You’ve got two curves, ht cos 2πt, t sin 2πt, t2 i
and ht cos 6πt, t sin 6πt, t2 i. Problem: Graph them and say what quadric surface
both live on.
Well, in both cases, x2 + y 2 = t2 (cos2 u + sin2 u) = t2 , where u is either 2πt
or 6πt and the trig squared stuff adds to 1 either way. Since z = t2 , this says
that x2 + y 2 = z for all points on both curves. That’s our quadric surface.
As to the curves themselves, they sort of spiral their way up the paraboloid.
One more rapidly, one more slowly. Here they are:
2
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