I have this surface, and I'm asked to find the tangent plane at the point

$(x,y,z) = (-\alpha, \alpha-1, -\alpha)\quad \alpha>0$

The surface is this one:

$(x,y,z)\in\mathbb{R}^3\quad\rvert\quad y+1 = \sqrt{ z^2 +2(x + \alpha)^2 }$

I did this way: I wrote this as

$\sqrt{ z^2 +2(x + \alpha)^2 } - y - 1 = 0$

and found the partial derivatives with respect to each direction:

The derivative with respect to $x,y,z$ turned out to be

$\frac{2(x+\alpha)}{ \sqrt{ z^2 +2(x + \alpha)^2} }\quad-1\quad\frac{z}{\sqrt{ z^2 +2(x + \alpha)^2}}$

Evaluating them at the given point I found $(0, -1, -1)$

To end with it the expression of the tangent plane should be

$\pi: 0(x + \alpha) -1[y - (\alpha - 1)] - 1(z +\alpha) = 0 \to \\\pi: y + z + 1 = 0$

Is this correct? (I'm surprised it doesn't depend on $\alpha$). Where can I find more exercises like this one?


That is correct.

Here are some practice problems.



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