# Parabolic bowl plane?

The surface with equation $z = x^2+ y^2$ is a parabolic bowl sitting on the origin. If I cut it with a plane perpendicular to the z-axis, I get a circle. If I cut it with a plane parallel to the z-axis, I get a parabola. Find the plane which is tangent to the surface at the point (1, 2, 5).

Now this looks like a fine problem in a multivariable calculus course, and indeed it is. But here I’m looking for an algebraic solution. There are actually lots of ways to do it, but here’s an idea: the tangent plane will have only one intersection with the bowl––at the point (1, 2, 5).

My professor always includes these questions at the end of his excercises, ones that are meant to challenge us, but they're always so intimidating. Can someone help me through the thought process to this question?

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Math 111 eh? Right there with ya. –  Lucas Oct 25 '12 at 5:39

I would start by working in the plane $y=2x$. Make a $w$ axis in this plane perpendicular to $z$ with unit vector $\frac 1{\sqrt 5}(1,2,0)$. In this plane, the equation is $z=w^2$. Now it is much easier to express the tangent in the $wz$ plane, then transform back to $xy$.
@Lucas: It is easier to cut the bowl with a plane that has all the activity. I left off the normalization of the unit in the $w$ direction, now fixed. The point of interest is $w=\sqrt 5, z=5$. If you find (through calculus or otherwise)that the slope is $2 \sqrt 5$ in the $wz$ plane at that point you can see the plane has to include the lines $z=0$ and $z=2 \sqrt 5 w-5$ –  Ross Millikan Oct 25 '12 at 13:02