# Intersection of Surfaces

1. Sep 18, 2013

### cp255

I need to find a parametric equation of the vector function which is the intersection of y = x2 and x2 + 4y2 + 4z2=16. I know the graph of the first equation is a parabola which stretches from negative infinity to infinity in the z direction. I also know that the second equation is that of an ellipsoid.

I have attempted to solve this by substituting x2 for y in the second equation which I think produces an ellipse. After this I am stuck.

2. Sep 18, 2013

### Dick

You want a PARAMETRIC form. You want to express x, y and z in terms of some parameter t. I would pick x=t to start. Now can you express the other variables in terms of t?

3. Sep 18, 2013

### Jolb

edit: why can't I delete my post? No replies yet. But yeah, take Dick's suggestion for a straightforward way to think about this problem.

4. Sep 18, 2013

### cp255

So I took Dick's approach and I got <t, t2, (1/2)(16 - t2 - 4t4)1/2>. The only problem is that this is half the solution. I know I could get the other half if I took the negative sqrt but for my class I need one equation. It's kind of like when you convert the circle x2 + y2 = 1 to a parametric form; you could get <t, (1-t2)1/2> which just gives you half the circle, or you can do it the better way and get <sin(t), cos(t)> which gives an entire circle.

5. Sep 18, 2013

### szynkasz

cp255, by substitution $y=x^2$ you got the ellipse equation. Complete the square and it will be easy to parametrize with $\cos t,\,\sin t$

6. Sep 18, 2013

### cp255

I know but then still I would have to define x as the square root of what y which only gives half the curve.