What I actually have to do is find the point on the surface which is closest to the point $P$, both of which are given below.
Function $z(x,y)=x^2+2y^2$
Point $P = (1,-1,1)$
This is what I have tried:
Consider a point $Q = (a,b,c)$ on this surface. Let this be the point that is closest to the point $P$. Since $Q \in z(x,y)$, we have that $c=a^2+2b^2$, so the point is $Q = (a,b,a^2+2b^2 )$. The vector $PQ$ is perpendicular to the surface at the point $(a,b,c)$.
$PQ=(a−1,b+1,a^2+2b^2−1)$
The distance:
$$ \begin{align*} D&=\sqrt{(a−1)^2+(b+1)^2+(a^2+2b^2−1)^2} \\ &=\sqrt{a^2−2a+1+b^2+2b+1+a^4+2a^2 b^2−a^2+2a^2 b^2+4b^4−2b^2−a^2−2b^2+1} \\ &=\sqrt{3+4a^2 b^2−2a+2b−a^2−3b^2+a^4+4b^4} \end{align*} $$
Minimize the distance by setting partial derivatives equal to 0:
$$ \begin{align*} D_a&=\frac{8b^2 a−2−2a+4a^3}{\sqrt{2(3+4a^2 b^2−2a+2b−a^2−3b^2+a^4+4b^4}} \\ &=\frac{4a^3+(8b^2−2)a−2}{\sqrt{2(3+4a^2 b^2−2a+2b−a^2−3b^2+a^4+4b^4}} \\ &\Rightarrow 0=4a^3+(8b^2−2)a−2 \end{align*} $$ $$ \begin{align*} D_b&=\frac{16b^3+(8a^2−6)b+2}{\sqrt{3+4a^2 b^2−2a+2b−a^2−3b^2+a^4+4b^4}} \\ &\Rightarrow 0=16b^3+(8a^2−6)b+2 \end{align*} $$ Basically this is where I've gotten to based on some of the suggestions I have seen for similar questions. I saw one method which looked nice here: (http://www.physicsforums.com/showthread.php?t=340772) but they lose me when they get the normal vector, which I have no idea where it came from. I have the actual answer (0.728, -0.573, 1.187) but cannot for the life in me figure out how to get there. I'm looking for errors in my arithmetic or another method if somebody has one (note: I do not know Lagrange Multipliers, which I have seen applied in some solutions, and they are not part of my curriculum, so it would not be useful to solve them this way)