points of intersection between cone and paraboloid

• Mar 27th 2014, 10:13 PM
junkwisch
points of intersection between cone and paraboloid
Hey everyone, I need to find the points of intersection between the cone $\displaystyle z=\sqrt(x^2+y^2)$and $\displaystyle z=2-x^2-y^2$

So, what I did was

$\displaystyle 2-x^2-y^2=\sqrt(x^2+y^2)$
$\displaystyle (2-x^2-y^2)(2-x^2-y^2)=x^2+y^2$
$\displaystyle x^4+y^4-5*x^2-5*y^2+2*x^2*y^2+4=0$

The problem is, I don't know what to do for now, can any one give me any tips?

Best Regards
Junks
• Mar 28th 2014, 12:52 AM
romsek
Re: points of intersection between cone and paraboloid
Quote:

Originally Posted by junkwisch
Hey everyone, I need to find the points of intersection between the cone $\displaystyle z=\sqrt(x^2+y^2)$and $\displaystyle z=2-x^2-y^2$

So, what I did was

$\displaystyle 2-x^2-y^2=\sqrt(x^2+y^2)$
$\displaystyle (2-x^2-y^2)(2-x^2-y^2)=x^2+y^2$
$\displaystyle x^4+y^4-5*x^2-5*y^2+2*x^2*y^2+4=0$

The problem is, I don't know what to do for now, can any one give me any tips?

Best Regards
Junks

$z = \sqrt{x^2+y^2}$

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

$z = 2 - z^2$

Spoiler:
$z^2 + z - 2 =0$

$(z+2)(z-1) = 0$

$(z=-2) \vee (z=1)$

The first equation for $z$ rules out $z=-2$ as a solution so the only solution is $z=1$ i.e.

$z=\sqrt{x^2+y^2}=1$ and this is a circle of radius 1 centered at (0,0,1) in the plane $z=1$.