# Tangent plane to surface

[SOLVED] Tangent plane to surface

## Homework Statement

Find an equation for the tangent plane to the surface z^2 = x^2 + 2y^2 at the point P = (1,2,3). Which other points on the surface have the same tangent plane?

2. The attempt at a solution

I find the derivatives:

fx = x/(2*sqrt(x^2 + 2y^2))
fy = 2y/(2*sqrt(x^2 + 2y^2))

The plane is given by:

z - f(1,2,3) = fx(1,2,3)(x-1) + fy(1,2,3)(y-2)

6z = 2x + 8y

So then, which other points on the surface have the same tangent plane?

I also have another question: Why does it suffice to write z = sqrt(x^2 + sy^2) instead of z = (+/-)sqrt(x^2 + sy^2) ?

HallsofIvy
Homework Helper
First,
[tex]f_y\ne \frac{2y}{2\sqrt{x^2+ 2y^2}}[/itex]
I suspect you forgot the "2" multiplying y2.

You are given that the point in question is (1, 2, 3). z= 3 which is positive. That's why you can use $z= \sqrt{x^2+ 2y^2}$ for that point. Of course, other points, having the same tangent plane, might have z negative. In order to determine what other points on the surface have the same tangent plane, you should first determine what other points on the surface are also on that plane. Once you have those, then look at the normal vectors.

By the way, for those of us who don't like square roots, it is simpler to write F(x,y,z)= x2+ 2y2- z2= 0, so the surface is a "level surface" for F. It's gradient, $\nabla F= 2x\vec{i}+ 4y\vec{j}- 2z\vec{k}$ is perpendicular to that surface. In particular, $\nabla F(1,2,3)= 2\vec{i}+ 8\vec{j}- 6\vec{k}$ gives exactly the same tangent plane.

Finally, please don't use "fx", "fy", etc. without first telling us what f is! I was able to guess that $f(x,y)= z= \sqrt{x^2+ 2y^2}$ but you should have said that.