Intersection of the hyperboloid with its tangent plane

Question

Find the intersection of the hyperboloid $x^2+y^2-3z^2=1$ with its tangent plane at the point $(2,0,1).$

I know the equation of the tangent plane at that point is $2x-3z=1$.
But isn't the intersection of the plane and the hyperboloid the point $(2,0,1)$ itself? Or did I miss something?

Answer 1

This tangent plane cuts the surface in two lines, as shows GeoGebra picture.

The system $$\begin{aligned} x^2+y^2-3z^2&=1\\ 2x-3z&=1 \end{aligned}$$ (plug $z$ from second into the first equation) gives $$3y^2-(x-2)^2=0,$$ which is union of the planes $$x-y\sqrt3=2 \quad \text{or} \quad x+y\sqrt3=2.\tag{1}$$

To obtain the lines (parametric or other kind of equations), use the tangent plane and one of the equations in $(1)$.

Answer 2

In ${\Bbb P}^3$ (Figure from 3264 and all that)

In ${\Bbb P}^3_{\Bbb R}:$ $x^2+y^2-3z^2-w^2=0$ (set $w=1$ to get back the affine surface in ${\Bbb A}^3_{\Bbb R}$) has signs ++-- so it has two rulings of real lines and the picture above is real.

Rewriting the affine equation about the point $(2,0,1)$: $$(x-2)^2+y^2-3(z-1)^2+4(x-2)-6(z-1)=0,$$ the tangent plane is the linear terms $4(x-2)-6(z-1)=0$ or as you say $2x-3z=1$.

The ideal of the system (again about $(2,0,1)$): $$\langle(x-2)^2+y^2-3(z-1)^2+4(x-2)-6(z-1),2(x-2)-3(z-1)\rangle$$ can be written: $$\langle 2(x-2)-3(z-1), 4y^2-3(z-1)^2\rangle$$ or $$\langle 2(x-2)-3(z-1), (2y-\sqrt{3}(z-1))(2y+\sqrt{3}(z-1))\rangle$$ or as the union of the two lines (intersection of the ideals) $$\langle 2(x-2)-3(z-1), 2y-\sqrt{3}(z-1))\rangle\cap \langle 2(x-2)-3(z-1), 2y+\sqrt{3}(z-1)\rangle.$$