The points closest to the origin

noob mathematician · #1 $Old$ November 26th, 2009, 01:17 AM

Question: The intersection of the plane $x+2y+3z=10$ and the surface $z=x^2+y^2$ is an ellipse E. Find the points on E that are closest to the origin.

What I did:

First I join the two equations together to form: $x+2y+3x^2+3y^2=10$

Using $d^2=x^2+y^2+z^2$ and treating the above equation as a constraint:
$2x=(1+6x)\lambda$
$2y=(2+6y)\lambda$
$x+2y+3x^2+3y^2=10$

I managed to get: $(x=2/3,y=4/3),(x=-1,y=-2)$

Subst back into the equation I have $z= 20/9 \text{ and } 5$ respectively.

So the points I conclude is $(2/3,4/3,20/9)$

Is this right? Or do I have to form two contraint equations instead?

HallsofIvy · #2 $Old$ November 26th, 2009, 03:38 AM

What you have done is perfectly valid.