Cutting a Paraboloid According to Specs

The equation of the paraboloid can be written as,

$x^2 + y^2 - z = 0$

So, if we define $r = (x, y, z)$ then we can write the paraboloid equation as,

$r^T Q r + b^T r = 0$

where,

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

Since the paraboloid is symmetrical about the z-axis, we can, without loss of generality, take the normal of the plane, to lie in the $xz$ -plane. If the angle it makes with the $z$ -axis is $\phi$ , then the unit normal vector can be taken as,

$\hat{n} = ( - \sin \phi , 0, \cos \phi )$

Orthogonal to the above defined normal and orthogonal to each other are these two unit vectors

$u_1 = ( \cos \phi, 0, \sin \phi )^T$ $u_2 = ( 0, 1, 0 )^T$

If we define the $3 \times 2$ matrix $V = [ u_1, u_2 ]$ , then any point on the plane, can be expressed as,

$r = r_0 + V u$

where $r_0$ is any fixed point on the plane, and for convenience we'll take it as the $z$ -intercept of the plane, that is,

we'll take $r_0 = (0, 0, z_0 )^T$ . (Remember, all vectors are considered column vectors). The $2 \times 1$ vector $u$ multiplying $V$ is the coordinate of the point $r$ with respect to the two axes $u_1$ and $u_2$ .

Substituting the expression for points on the plane into the equation of the paraboloid gives us

$( r_0 + V u)^T Q (r_0 + V u) + b^T (r_0 + V u) = 0$

Expanding,

$u^T V^T Q V u + 2 u^T V^T Q r_0 + {r_0}^T Q r_0 + b^T r_0 + u^T V^T b = 0$

Now, we'll evaluate each of these terms.
$\begin{aligned} V^T Q V &= \begin{bmatrix} \cos \phi && 0 && \sin \phi \\ 0 && 1 && 0 \end{bmatrix} \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 0 \end{bmatrix} \begin{bmatrix} \cos \phi && 0 \\ 0 && 1 \\ \sin \phi && 0 \end{bmatrix} \\ &=\begin{bmatrix} \cos \phi && 0 && 0 \\ 0 && 1 && 0 \end{bmatrix} \begin{bmatrix} \cos \phi && 0 \\ 0 && 1 \\ \sin \phi && 0 \end{bmatrix} \\ &= \begin{bmatrix} \cos^2 \phi && 0 \\ 0 && 1 \end{bmatrix} \end{aligned}$

$V^T Q r_0 = \begin{bmatrix} \cos \phi && 0 && 0 \\ 0 && 1 && 0 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ z_0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix}$

${r_0}^T Q r_0 = \begin{bmatrix} 0 && 0 && z_0 \end{bmatrix} \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 0 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ z_0 \end{bmatrix} = 0$

$b^T r_0 = \begin{bmatrix} 0 && 0 && -1 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ z_0 \end{bmatrix} = - z_0$

$V^T b = \begin{bmatrix} \cos \phi && 0 && \sin \phi \\ 0 && 1 && 0 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} -\sin \phi \\ 0 \end{bmatrix}$

So the equation of the cut is as follows:

$u^T (V^T Q V) u + u^T V^T b = z_0$

We next introduce the vector $u_0$ defined as follows,

$u_0 = -\dfrac{1}{2} (V^T Q V)^{-1} V^T b = -\dfrac{1}{2} \begin{bmatrix} \frac{1}{\cos^2 \phi} && 0 \\ 0 && 1 \end{bmatrix} \begin{bmatrix} -\sin \phi \\ 0 \end{bmatrix} = \dfrac{1}{2} \begin{bmatrix} \frac{\sin \phi}{\cos^2 \phi} \\ 0 \end{bmatrix}$

Then, we can write now write the equation of the cut as,

$( u - u_0)^T V^T Q V ( u - u_0) = z_0 + {u_0}^T (V^T Q V) u_0$

Evaluating the rightmost term, results in,

$( u - u_0)^T (V^T Q V) ( u - u_0) = z_0 + \dfrac{1}{4} \tan^2 \phi$

Dividing by the right hand side, we obtain

$(u - u_0)^T Q_1 (u - u_0) = 1$

where

$Q_1 = \begin{bmatrix} \dfrac{ \cos^2 \phi }{ z_0 + \dfrac{1}{4} \tan^2 \phi } && 0 \\ 0 && \dfrac{1}{ z_0 + \dfrac{1}{4} \tan^2 \phi } \end{bmatrix}$

Now we can compare this form to the equation of an ellipse with semi-major and semi-minor axes $a$ and $b$ which is given by:

$w^T D w = 1$ where

$D = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && \dfrac{1}{b^2} \end{bmatrix}$

Comparing the two matrices $Q_1$ and $D$ , we deduce that,

$\dfrac{b}{a} = \cos \phi$ and that $b^2 = z_0 + \dfrac{1}{4} \tan^2 \phi$

Therefore,

$\cos \phi = \dfrac{3}{5} = 0.6$ and $z_0 = 9 - \dfrac{1}{4} \dfrac{16}{9} = 9 - \dfrac{4}{9} = \dfrac{77}{9}$