Find the normal curvature of the hyperboloid$$\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$$ at the vertex of the hyperboloid.
I know the rest of the work, but how to find the vertex of such hyperboloid? The problem says it is two-sheets-hyperboloid (or two surface). Should I find the minimum value that $x$ can take (I guess this from the picture attached) or what?
2 Answers
$\begingroup$I've never heard about vertex of a hyperboloid, but I suppose that, here, it means the points $(\pm a,0,0)$. In fact, these points belong to the hyperboloid, and if $(x,y,z)$ is such that $|x|<a$, then$$\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}\leqslant\frac{x^2}{a^2}<1$$and therefore $(x,y,z)$ does not belong to hyperbolod.
$\endgroup$ 4 $\begingroup$As the equation of hyperboloid is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1$. It is a hyperboloid of two sheets with x-axis passing through them. At the points that x-axis intersects both sheets of the hyperboloid is being referred to vertices here. So we plug in $y = z = 0$ and that gives us $x = \pm a$.
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