Equation of lines of intersection of tangent plane to a one sheet hyperboloid

Question

I want to find the general equation of the two lines of intersection of a one sheet hyperboloid to its tangent plane for the function

$F(x,y,z)=x^2+y^2-z^2=1$

at

$(x_0,y_0,z_0)$,

The equation of the tangent plane is

$x_0x+y_0y-z_0z=1$

and I've tried substituting this into the hyperboloid equation and I haven't been able to do anything the following mess.

$(z_0^2-x_0^2)x^2+(z_0^2-y_0^2)y^2+2x_0x+2y_0y-2x_0y_0xy-z_0^2-1=0$

I have been provided answers but I don't know how to get there and it's been bugging me for about three days now. How do I get the equation of line of intersection from here?

Answer 1

The parameterization of the given hyperboloid is

$(x, y, z) = ( \sec t \cos s , \sec t \sin s , \tan t ) $

Set $P_0=(x_0, y_0, z_0) = ( \sec t_0 \cos s_0 , \sec t_0 \sin s_0 , \tan t_0 ) $ and find $t_0$ and $s_0$.

Next, the normal to the tangent plane is

$ N = (x_0, y_0, -z_0) = ( \sec t_0 \cos s_0, \sec t_0 \sin s_0, - \tan t_0 )$

Construct two unit vectors that are normal to $N$, and normal to each other, these can be chosen as

$ u_1 = ( - \sin s_0, \cos s_0 , 0 ) $

and

$ u_2 = ( \cos s_0 \tan t_0 , \sin s_0 \tan t_0, \sec t_0 ) / \sqrt{2 \tan^2 t_0 + 1} $

Now points on the plane can be expressed as

$ P = P_0 + \alpha u_1 + \beta u_2 = P_0 + U X $

where $U = [u1, u2] $ and $ X = [\alpha, \beta]^T $, while the equation of our hyperboloid is

$ r^T Q r = 1 $

with $Q = \text{diag}[ 1, 1, -1 ] $

Plugging in the expressions for $P$ into the hyperboloid equation, results in,

$ (P + U X)^T Q (P + U X) = 1 $

Expanding,

$ P^T Q P + 2 X^T U^T Q P + X^T U^T Q U X = 1 $

Since $P$ is on the hyperboloid , then $P^T Q P = 1 $, thus the equation becomes,

$ X^T U^T Q U X + 2 X^T U^T Q P = 0 $

We now need to compute $ U^T Q U $ and $U^T Q P $, the reader can verify that,

$U^T Q U = \begin{bmatrix} 1 && 0 \\ 0 && \dfrac{-1}{2 \tan^2 t_0 + 1 } \end{bmatrix}$

and that,

$U^T Q P = \mathbf{0} $

Hence, the equation becomes,

$ X^T U^T Q U X = 0 $

Substituting $U^T Q U$ and $X$, yields,

$ \alpha^2 - \dfrac{1}{2 \tan^2 t_0 + 1 } \beta^2 = 0 $

Thus

$ \beta = \left( \pm \sqrt{2 \tan^2 t_0 + 1} \right) \alpha $

i.e.

$ X = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = t \begin{bmatrix} 1 \\ \pm \sqrt{2 \tan^2 t_0 + 1 } \end{bmatrix} $

where $t \in \mathbb{R} $ is arbitrary. This is an equation of a line, and the $\pm$ indicates that we have two lines.

Answer 2

Here is a simple approach.

I begin by adapting the information I gave in the following answer some time ago.

The 2 families of skew lines $L_a$ and $L'_b$ generating hyperboloid with one sheet $(H)$ can be retrieved, starting from its equation

$$x^2+y^2-z^2=1 \ \ \ \iff \ \ \ (y-z)(y+z)=(1-x)(1+x),\tag{1}$$

in the following natural way:

$$\text{Lines} \ L_a : \ \begin{cases}y-z&=&a(1-x)\\y+z&=&\dfrac{1}{a}(1+x)\end{cases}\tag{2}$$

$$\text{Lines} \ L'_b : \ \begin{cases}y-z&=&b(1+x)\\y+z&=&\dfrac{1}{b}(1-x)\end{cases}\tag{3}$$

for any non-zero real number $a$ or $b$.

Indeed: by multiplication of its 2 equations, (2) $\implies$ (1) ; implication of equations meaning inclusion of corresponding geometric entities ($\forall a, L_a \subset H$) as desired. For the same reason, $\forall b, L'_b \subset H$.

Therefore, for a given point $(x_0,y_0,z_0)$, you just have to find the values of coefficients $a$ and $b$, which is straightforward.

Consider the case of $a$. From the first equation in (2), one gets:

$$a=\frac{y_0-z_0}{1-x_0}=\frac{y_0 \pm \sqrt{x_0^2+y_0^2-1}}{1-x_0}\tag{4}$$

which is valid under the condition that $x_0 \ne 1$. If $x_0=1$, get $a$ instead from the second equation in (2).

Do the same for $b$ and plug these expressions into (2), resp. (3).