Tangent plane at a point of hyperboloid of one sheet

Question

The tangent plane at a point of a hyperboloid meets the hyperboloid in the two generators through that point. However, a tangent plane is defined as the locus of all tangent lines at a point on the hyperboloid. Now, since generators lie completely on the surface they can't be tangent lines to the hyperboloid. Then how is it possible that the tangent plane contains the generators that intersect at that point?

Answer 1

The plane without the two intersecting lines is not a plane as such, so taking the closure is a very natural thing to do, and I suspect that's what we do do in algebraic geometry.

Using the definition of tangent plane from algebraic geometry, there is no confusion.

Take an example: $x^2+y^2-z^2-1=0$ at the point $(a,b,c)$ lying on the surface (i.e. satisfying $a^2+b^2-c^2-1=0$) has tangent plane $2a(x-a)+2b(y-b)-2c(z-c)=0$ (by rewriting the equation as $(x-a)^2+(y-b)^2-(z-c)^2+2a(x-a)+2b(y-b)-2c(z-c)+a^2+b^2-c^2-1=0$ and taking the tangent cone, which is just the tangent plane in this case of the surface being smooth. You get the tangent cone by keeping just the linear terms locally around the point) or $ax+by-cz=1$ if you want to zoom out from the point $(a,b,c)$. To be specific and take $(a,b,c)=(1,1,1),$ we get $x+y-z=1.$

And intersecting we get the two lines $(1,1,1)+t(-\frac{\sqrt2}{2}\pm\frac{\sqrt2}{2},\frac{\sqrt2}{2}\pm\frac{\sqrt2}{2},\pm\sqrt2).$