I can't see a reason to reject one solution. Parametized surface Ok so the question is to consider the parametrized surface
x = $u^2 - v^2$,    y = u + v, z= $u^2 + 4v$
It asks for me to find a vector normal to the surface in terms of u and v which is no problem.
(4- 2u, -8u - 4uv, 2u + 2v)
Which I know is correct: then it asks for the tangent plane at ($\frac{-1}{4},\frac{1}{2},2)$
Which I thought would be no problem just dot product of the normal vector at this point with the whole (x - x0) but Im having troubles getting a single u and v I get $\frac{1}{2}$ or $\frac{-9}{2}$ I just don't know anything else to reject one of them it gives the answer as 4x+z =1 which is fantastic I could figure out both and just reject the one that isnt the answer but it doesnt help me get ready for exams.
 A: You have (at your point) $9/4 = 2-(-1/4) = z-x = (u^2+4v)-(u^2-v^2) = v^2+4v$. So $v = -9/2$ or $1/2$. Next, $u=y-v = 1/2-(-9/2)=5$ or $1/2-1/2=0$. So either $u=5$ and $v=-9/2$ or $u=0$ and $v=1/2$.
If $u=5$ and $v=-9/2$, then $x=5^2-(-9/2)^2=25-36/4 = 16$ (which is wrong).
If $u=0$ and $v=1/2$, then $x=0^2-(1/2)^2=-1/4$, $y=0+1/2=1/2$, and $z=0^2+4(1/2)=2$ (which is correct).
So $u=0$ and $v=1/2$ is the (only) solution.
Explanation: When initially finding $u$ and $v$, we used the following equations: $z-x=v$ and $u=y-v$. This amounts to using a combination of the first and third equations ($x=u^2-v^2$ and $z=u^2+4v$) and the second equation ($y=u+v$). We then arrive at our 2 solutions. Thus this (smaller) system of equations does in fact have 2 solutions. But when we plug these back into our (bigger) system of 3 equations, we find out that one of them is a solution of all 3 equations and the other isn't. This often happens when we solve systems of (non-linear) equations. While searching for a solution we pare down the number of equations (eliminating and substituting). Sometimes in the course of throwing out (seemingly irrelevant) information, we allow "new solutions" to creep in (which aren't actually solutions of the original equations).
