Find equation of tangential plane at a given point of a surface

Question

I have a parametric, regular surface $$s(u,v)=(\cosh(u)\cos(v),\cosh(u)\sin(v),\sinh(u)),$$ $(u,v)\in\mathbb{R}^2$. I need to find tangent plane in point $Q=s(1,0)$. I used the formula for calculation the plane but the problem is I can't get an explicit equation.

I got $$(x, y, z) = (\cosh(1)+u\sinh(1)-\sinh(1),v\cosh(1), \sinh(1)+u\cosh(1)-\cosh(1)).$$

Have I missed something or is there a better way. I know that this surface is a surface of one sheet hyperboloid but I can't use the standard formula.

Answer 1

You were doing fine until the last step, when you used a template for a parametric equation of the tangent plane, i.e., a linear approximation to $s$ at $(1,0)$. To get an implicit Cartesian equation, you can either eliminate $u$ and $v$ from the result (note that they’re not the same $u$ and $v$ as in the parameterization of the surface), or use the point-normal form of plane equation in the first place. To do the latter, you need a normal vector to the surface at $s(1,0)$, which is just $s_u(1,0)\times s_v(1,0) = (-\cosh^2(1),0,\cosh(1)\sinh(1))$, which you can divide by $-\cosh(1)$ to keep the expressions simpler. The implicit equation of the tangent plane at $s(1,0)$ is then $\mathbf n\cdot\mathbf x = \mathbf n\cdot s(1,0)$, or $$x\cosh(1)-y\sinh(1)=\cosh^2(1)-\sinh^2(1) = 1.$$

Answer 2

Write the surface as

$$x^2+y^2-z^2=1$$

and the point $s(1,0)$ corresponds to $\vec{Q} = (\cosh(1),0,\sinh(1))$. The vector normal to the surface at the point $(x,y,z)$ is given by $(z_x',z_y',-1$), or

$$\vec{n} = (\coth(1),0,-1)$$

Let $\vec{P}=(x,y,z)$ be the points on the tangent plane, which has to satisfy $\vec{n}\cdot\vec{QP}=0$, or,

$$(\coth(1),0,-1)\cdot(x-\cosh(1),y,z-\sinh(1))=0$$

Thus, the equation of the tangent plane is

$$\cosh(1)x-\sinh(1)z=1$$