Intersection with paraboloid This problem feels really easy but I've been having a really hard time with it. I'm given an equation of a paraboloid $z=x^2+4y^2$ and told that an unknown plane, perpendicular to the $xy$ plane has a point $(2,1,8)$ in common with the paraboloid. The intersection between the plane and the paraboloid is a parabola with slope $0$ at the given point.
I'm told to find the equation of the plane. I've tried using the gradient vector but I found out that my approach is wrong. I tried to explicitly find the intersection between a plane with an equation $y=ax+d$ and the paraboloid equation, then differentiate it once to find out what $a$ and $d$ are so that the slope in $(2,1,8)$ is $0$, looking at the graphs in Mathematica it seems that I've got it wrong with both approaches. Looking for any suggestions on this, I'm really lost.
Edit: some information on the gradient approach. I calculated $\nabla z(x,y)=(2x,8y)$, then substituted $x$ and $y$ for $2$ and $1$ respectively. This should be perpendicular to the level curve $8=x^2+4y^2$, if I'm thinking correctly. Therefore, I can define a plane using $\nabla z(2,1)=(4,8)$ and using the fact that we know the plane is orthogonal to $xy$, therefore we use $(4,8,0)$ as the normal vector to the plane. So, by my chain of thought, the plane equation should be $4x+8y+d=0$, substituting $x$ and $y$ for $2$ and $1$ we get $d=-16$. Unless I messed up my Mathematica plot lots of times, this isn't right...
 A: If the slope of the parabolic intersection at the point $(2,1,8)$ is zero, then that point is the vertex of the parabola. So the plane being sought will be tangent to the level curve $x^2+4y^2=8$ at the point $(2,1,0)$. That line has slope $m=-\dfrac{1}{2}$. 
So the intersection of the plane being sought and the $xy$-plane will be the line with slope $m=-\dfrac{1}{2}$ and containing $(2,1,0)$, namely $y=-\dfrac{1}{2}x+2$ which is also the equation of the plane.

A: Your approach is correct. The plane is $y = ax + d$, so the intersection is a parabola $z = x^2 + 4(ax + d)^2$. The point $(2,1,8)$ gives two restrictions on $a$ and $d$: $1 = 2a + d$ (because point lies on the plane) and $8 = 2^2 + 4(2a + d)^2$ (because point lies on a paraboloid). Notice that they are identical, and conclude that $d = 1-2a$. The parabola equation becomes
$$z = x^2 + 4(ax + 1 - 2a)^2$$
Differentiate wrt $x$. The derivative must vanish at $x = 2$. This leads to a quadratic equation for $a$. Solve it.
A: The normal to the paraboloid and $(2,1,8) = (4,8,-1)$
$4x + 8y - z = 8$ is a plane tangent to the paraboliod at the point (1,2,8)
I say that our target plane intersects the plane described above and forms a line that is parallel to the $xy$ plane.  This way the parabola formed by the intersection of our target plane and the paraboliod can be tangent to this line and at a minimum.
$x + 2y = 4$
lets check.
substitute
$x = 4 - 2y$
and hopefully we get a minimum at (1,8)
$z = (4-2y)^2 + 4y^2\\
z = 8 y^2 - 16 y + 16$
looks good.
