# Homework Help: Intersection point

1. Apr 26, 2010

1. The problem statement, all variables and given/known data

Does the surface $z = x^2 + y^2 - 4$ * intersects the $yz$-plane? If so, find the equation of the curve and write down the points of intersection.

3. The attempt at a solution
$yz-$plane, so $x=0$

1) * becomes $z = y^2 - 4$ and this is the equation of the curve that intersects the $yz$-plane.
2) So the intersection points are $(0,-2,0)$ and $(0,2,0)$

Are 1) and 2) correct? Thanks in advance!

2. Apr 26, 2010

### Je m'appelle

Shouldn't there be a third intersection point?

You found the intersection points of the curve on the y-axis, what about the z-axis?

3. Apr 26, 2010

Is it $(0,0,-4)$ ? But I am slightly confused, doesn't the whole curve intersect yz-plane and so there are infinitely many points of intersection?

Thanks.

4. Apr 26, 2010

### lanedance

the graph of $z = x^2 + y^2 - 4$, is a parabaloid, rotationally symmetric about the z axis

the intersection of the surface with the yz plane (x=0) is the parabola $z = y^2 - 4$ which has, as you say, infinitely many points.

I'm not too sure what the last part is asking for, maybe it is where the curve intersects the y & z axis....

5. Apr 26, 2010

I wanted to do the question myself so I only show a part of the question.
The original question is:
If the surface intersects the $xy, xz$, and $yz$ coordinate planes, find the equation of each of the parabolic boundary curves. Where do the curves intersect with the three coordinate axes?

I thought if the surface intersects the three coordinate planes, then the points of intersection must be $(0,0,0)$?

Thanks.

Last edited: Apr 26, 2010
6. Apr 26, 2010

### lanedance

nope, that is not a point on the surface, and so none of teh curves pass through that point

the easiest way to work out what i going on is to try & draw a picture, so i hope you'e attempted that. This problem is essentially showing you agood way to draw simple 3D surfaces, by drawing the intersection on each coordinate plane.

you should get 2 parabolas & and a circle

if you think about the shape of the surface and curves, the curve constrained by x = 0, will never intersect the x axis. So each point of intersection with axis will have 2 curves passing through it

due to the symmetry (which can save you some time) there will be one point of intersection on the z axis and 2 each for the x & y...

7. Apr 26, 2010

I have drawn the graph, it's a paraboloid. But I don't understand about the 2 parabolas and a circle.

I understand that the level curves are parabola and circle, but why the intersecting ones are only the specific 2 parabolas and 1 circle that you mentioned?

Thanks.

8. Apr 26, 2010

### lanedance

ok, so what is the equation for each curve...

9. Apr 26, 2010

$z = x^2 - 4$
$z = y^2 - 4$
$x^2 + y^2 = 4$

Are these correct?

10. Apr 26, 2010

### lanedance

yep so what do the forms of those equations look like?

11. Apr 26, 2010

The first two are parabolas. The last one is circle.
So are the intersection points
1) $(-2,0,0),(2,0,0)$ for the first curve.
2) $(0,-2,0),(0,2,0)$ for the second curve.
3) $(0,0,-4)$ for the third curve mentioned above?

Thanks.

12. Apr 26, 2010

### lanedance

teh points look ok, but they correspond to an axis, not a specific curve, each will have 2 curves passing through it

13. Apr 26, 2010

I am confused. Could you show me the equation for one the curves so I can try to figure what you mean? Didn't I already show the equation of the 3 curves? Are they different from the ones that you just mentioned? Thanks.

Last edited: Apr 26, 2010
14. Apr 26, 2010

### lanedance

you've got it all, so pulling it together from your posts (though check it as i did pretty quick):
curve 1 - yz plane (x=0)
$z = y^2 - 4$
curve 2 - xz plane (y=0)
$z = x^2 - 4$
curve 3 - xy plane (z=0)
$x^2 + y^2 = 4$

axes intersection points
x axis - (-2, 0, 0) and (2,0,0) by curves 2&3,
y axis - (0 ,-2, 0) and (0,2,0) by curves 3&1,
z axis - (0 , 0, 4) by curves 1&2

15. Apr 26, 2010