How to find the intersection between a plane and a conic?

Question

Find k such that the intersection of the plane kx+y=1 and the two-sheet hyperboloid x²−y²−z²=1 is an: a) ellipse b) hyperbola.

Answer 1

Define two mutually orthogonal unit vectors $u_1$ and $u_2$ that are both orthogonal to the normal of the plane. The plane is $kx + y = 1 $, so we can take

$\mathbf{v_1} = [1, -k, 0]^T /\sqrt{1 + k^2} , \mathbf{v_2} = [0, 0, 1]^T$

Define $\mathbf{r}=[x,y,z]^T$ , then the parametric equation of the plane is

$\mathbf{r} = \mathbf{r_0} + u_1 \mathbf{v_1} + u_2 \mathbf{v_2}$

for arbitrary scalars $u_1 , u_2 $. The above equation can be written compactly as

$\mathbf{r} = \mathbf{r_0} + \mathbf{V u} \hspace{12pt}(1) $

where $ \mathbf{V} = [\mathbf{v_1}, \mathbf{v_2} ] $ and $ u = [u_1, u_2]^T $

and $\mathbf{r_0} $ is any point on the plane and can be taken as $[0, 1, 0]^T$ On the other hand the equation of the given quadric is

$ \mathbf{r}^T \mathbf{Q r} = 1 \hspace{12pt} (2)$

where

$Q = \begin{bmatrix} 1 && 0 && 0 \\ 0 && -1 && 0 \\ 0 && 0 && - 1 \end{bmatrix}$

Substituting (1) into (2), we obtain,

$ (\mathbf{r_0} + \mathbf{V u})^T Q (\mathbf{r_0} + \mathbf{V u}) = 1$

Expanding,

$ \mathbf{u}^T \mathbf{V}^T Q \mathbf{V u} + 2 \mathbf{u}^T \mathbf{V}^T \mathbf{Q r_0} + \mathbf{r_0}^T \mathbf{Q} \mathbf{r_0} = 1 \hspace{12pt}(3) $

At the point, it is useful to plug in $\mathbf{Q}$, $\mathbf{V}$ and $\mathbf{r_0}$ and evaluate the terms of (3).

We have,

$\mathbf{V}^T \mathbf{QV} = \begin{bmatrix} \dfrac{1 - k^2}{1 + k^2} && 0 \\ 0 && - 1 \end{bmatrix} $

$ \mathbf{V}^T \mathbf{Q r_0} = \begin{bmatrix} -\dfrac{k}{\sqrt{1 + k^2}} \\ 0 \end{bmatrix} $

$ \mathbf{r_0}^T \mathbf{Q r_0} = -1 $

Hence the equation in $u_1$ and $u_2$ is

$\dfrac{(1 - k^2)}{(1 + k^2)} u_1^2 - u_2^2 - 2 \dfrac{k}{\sqrt{1 + k^2}} u_1 = 2 $

If $k = \pm 1$ , we get a parabola, so we can assume that $k \ne \pm 1 $.

Multiplying through by ${1 + k^2}$

$ (1 - k^2) u_1^2 - (1 + k^2) u_2^2 - 2 k \sqrt{1 + k^2} u_1 = 2 (1 + k^2)$

Completing the square,

$ (1 - k^2) ( u_1 - \dfrac{k \sqrt{1 + k^2}}{1 - k^2} )^2 - (1 + k^2) u_2^2 = 2 (1 + k^2) + \dfrac{k^2 (1 + k^2)}{1 - k^2} \hspace{12pt} (4) $

From (4) , it follows that if $ -1 < k < 1 $ then we get a hyperbola, and if $| k | > 1 $, we need $2 + \dfrac{k^2}{1 - k^2} \lt 0 $ to get an ellipse, i.e. $ 2(1 - k^2) + k^2 \gt 0 $ which simplifies to $k^2 \lt 2$ otherwise there is no intersection between the plane and hyperboloid.

To summarize:

• If $|k| \lt 1 $, the intersection is a hyperbola.
• If $k \in (-\sqrt{2}, -1) \cup (1, \sqrt{2})$ , then the intersection is an ellipse.
• If $k= \pm 1$, the intersection is a parabola.
• If $k = \pm \sqrt{2}$ , the intersection is a single point (tangent point).
• For all other values of $k$ there is no intersection between the plane and the hyperboloid.