Splitting a Parabolic Sheet

The equation of the given parabolic sheet is

$0.25 y^2 - z = 0$

Define

$r = [x, y, z]^T$

Then the equation of the parabolic sheet can be written as

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

where

$Q = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0.25 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ and $b = \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}$

On the other hand, a point in the cutting plane can be written in parametric form as

$r = r_0 + V u$

where $r_0$ is the position vector of any point on the plane, and can be taken as $(2, 0, 0)$ , and matrix $V$ is composed of two columns of two vectors $v_1$ and $v_2$ that are perpendicular to the normal vector to the plane. Further we will take $v_1$ and $v_2$ to be perpendicular to each other and to be of unit length.

The normal to the given plane is

$\hat{n} = [\sin \theta \cos \phi , \sin \theta \sin \phi , \cos \theta]^T$

So we can take,

$v_1 = [- \sin \phi, \cos \phi, 0 ]$

and

$v_2 = \hat{n} \times v_1 = [ -\cos \theta \cos \phi, -\cos \theta \sin \phi, \sin \theta]$

Now we can substitute the equation of the plane into the equation of the parabolic sheet

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

Expanding

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

Rearranging

$u^T V^T Q V u + u^T ( 2 V^T Q r_0 + V^T b ) = -r_0^T Q r_0 - b^T r_0$

Next, we diagonalize the matrix $V^T Q V$ , so that it can written as $V^T Q V = R D R^T$ , with matrix $R$ being an orthonormal matrix of unit eigenvectors.

And the above equation becomes,

$u^T R D R^T u + u^T ( 2 V^T Q r_0 + V^T b ) = -r_0^T Q r_0 - b^T r_0$

By choosing the columns of R appropriately, we can make the first entry on the diagonal of the diagonal matrix $D$ nonzero, and the second entry zero.

Now define the vector $w = R^T u$ , then

$w^T D w + w^T R^T ( 2 V^T Q r_0 + V^T b ) = - r_0^T Q r_0 - b^T r_0$

This equation is of the form,

$D_{11} w_1^2 + \alpha w_1 + \beta w_2 = \gamma$

So, by completing the square, it becomes,

$D_{11} (w_1 + \dfrac{\alpha}{2 D_{11}} )^2 + \beta w_2 = \gamma + \dfrac{\alpha^2}{4 D_{11}}$

This is an equation of a parabola, to put it the form given in the problem, we just shift the origin of the coordinate system to the vertex.

Define $z = w + w_0$ , where $w_0^T =[ \dfrac{\alpha}{2 D_{11}} , - \dfrac{1}{\beta} ( \gamma + \dfrac{\alpha^2}{4 D_{11}} ) ]$

Then,

$D_{11} z_1^2 + \beta z_2 = 0$

Therefore, the required parameter $a$ is given by $a =- \dfrac{ D_{11}} {\beta}$ .

Evaluating all of the above, yields

$R = \begin{bmatrix} 0.99613611 & 0.087822836 \\ -0.087822836 & 0.99613611 \end{bmatrix}$

$D = \begin{bmatrix} 0.244346183 & 0 \\ 0 & 0 \end{bmatrix}$

$\alpha = 0.076056807$

$\beta = - 0.862679177$

$\gamma = 0$

Therefore,

$a =- \dfrac{0.244346183}{(-0.862679177)} = 0.2832$