As the title states, I am struggling to parametrise the hyperbola resulting from intersection of a double cone and a plane. The equation of the cone is given as $z^2=x^2+y^2$ and the equation of the plane is $x+y=1$.
EDIT: I am stuck once I have some expressions for z.
Any help would be greatly appreciated.
EDIT: I have gotten as far as $z^2=1-2xy$ and $z^2=2x^2-2x+1=(x-\frac{1+i}2)(x-\frac{1-i}2)$.
Thank you, Robert.
The algebraic equation of the cone is
$ r^T Q r = 0 $
where $r =[x, y, z]^T $ and
$Q = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && - 1\end{bmatrix} $
The algebraic equation of the plane is
$ n^T (r - r_0) = 0 $
where $n = [1, 1, 0] $ and $r_0$ is any point on the plane, so I'll take it as $r_0= [1, 0, 0]$
From this algebraic equation, we can generate the vector equation of the plane by finding two mutually orthogonal unit vectors that span the plane, an obvious choice is
$u_1 = [0, 0, 1]^T $ and $u_2 = [1, -1, 0]^T / \sqrt{2} $
So now the vector equation of the plane is
$ r = r_0 + u u_1 + v u_2 = r_0 + U w $
where $U = [u1, u2] $ and $w = [u, v]^T $
Plug this vector expression for $r$ into the algebraic equation of the cone, you get,
$ (r_0 + U w )^T Q (r_0 + U w) = 0 $
Everything here is constant except the two entries of vector $w$.
Expand the above expression,
$ w^T U^T Q U w + 2 w^T U^T Q r_0 + r_0^T Q r_0 = 0 $
Direct evaluation gives us,
$A = U^T Q U = \begin{bmatrix} -1 && 0 \\ 0 && 1 \end{bmatrix} $
$b = U^T Q r_0 = \begin{bmatrix} 0 \\ \dfrac{1}{\sqrt{2}} \end{bmatrix} $
$c = r_0^T Q r_0 = 1 $
So our algebraic equation of the intersection is now
$ w^T A w + 2 w^T b + c = 0 $
To find the center of this conic, use the following formula
$ w_0 = - \dfrac{1}{2} A^{-1}(2 b ) = - A^{-1} b = \begin{bmatrix} 0 \\ - \dfrac{1}{\sqrt{2}} \end{bmatrix} $
And with this center, we can write the above equation as
$ (w - w_0)^T A (w - w_0) + c - w_0^T A w_0 = 0 $
Now $c - w_0^T A w_0 = 1 - \dfrac{1}{2} = \dfrac{1}{2} $
And now we have
$ (w - w_0)^T A (w - w_0) = -\dfrac{1}{2} $
Dividing through by $(-\dfrac{1}{2}) $ results in
$ (w - w_0)^T E (w - w_0 ) = 1 $
where $E = -\dfrac{1}{2} A = \begin{bmatrix} 2 && 0 \\ 0 && - 2 \end{bmatrix} $
Matrix E is already diagonal, so now define the vector $v = w - w_0$ then
$ 2 v_1^2 - 2 v_2^2 = 1 \hspace{25pt} (*)$
And this the equation of a hyperbola that opens in both directions along the $v_1$ axis which is $ w_1$ axis (i.e. along $u_1$ ).
The standard parameterization for the hyperbola $(*)$ is
$ v_1 = \dfrac{1}{\sqrt{2}} \sec t, \hspace{25pt} v_2 = \dfrac{1}{\sqrt{2}} \tan t \hspace{25pt} t \in \mathbb{R} $
Recalling the relation between $v$ and $w$, we deduce that
$ w_1 = v_1 = \dfrac{1}{\sqrt{2}} \sec t $ , and
$ w_2 = -\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} \tan t $
Finally, the position vector $r = r_0 + U w $, so $r$ is given by
$ r = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 && \dfrac{1}{\sqrt{2}} \\ 0 && -\dfrac{1}{\sqrt{2}} \\ 1 && 0 \end{bmatrix} \begin{bmatrix}\dfrac{1}{\sqrt{2}} \sec t \\ -\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} \tan t \end{bmatrix}=\begin{bmatrix} \dfrac{1 + \tan t }{2} \\ \dfrac{1 - \tan t }{2} \\ \dfrac{1}{\sqrt{2}} \sec t \end{bmatrix}$
And this is a parameterization of the hyperbola of intersection.