Oblique projection of an ellipse on the $xy$ plane - Axes

The vector passing through a point on the edge of the ellipse is defined by the equation $(x, y, z) = (-1, -2, -5)u + \mathbf{p}(t)$ .

For the shadow on the $xy$ -plane, $z = -5u + 10 - 3 \cos t + 9 \sin t = 0$ , which rearranges to $u = 2 - \frac{3}{5} \cos t + \frac{9}{5} \sin t$ .

Substituting $u$ back into the vector equation and simplifying gives $(x, y) = (3 + \frac{23}{5} \cos t + \frac{6}{5} \sin t, \frac{21}{5} \cos t + \frac{7}{5} \sin t)$ , an ellipse on the $xy$ -plane that has a center at $(3, 0)$ .

The square of the distance between a point on the edge of the ellipse and its center is $d^2 = (3 + \frac{23}{5} \cos t + \frac{6}{5} \sin t - 3)^2 + (\frac{21}{5} \cos t + \frac{7}{5} \sin t))^2$ which rearranges and simplifies to $d^2 = \frac{57}{5} \sin 2t + \frac{177}{10} \cos 2t + \frac{211}{10}$ .

By implicit differentiation, $2d \frac{dd}{dt} = \frac{114}{5} \cos 2t - \frac{177}{5} \sin 2t$ . The semi-major and semi-minor axes will be when $\frac{dd}{dt} = 0$ , which after substituting solves to $t = \frac{1}{2}n \pi + \frac{1}{2} \tan^{-1} \frac{38}{59}$ for any integer $n$ .

Substituting $t$ back into $d^2$ gives two values for $d$ , which correspond to the semi-major and semi-minor axes, $a = \sqrt{\frac{211}{10} + \frac{3\sqrt{197}}{2}}$ and $b = \sqrt{\frac{211}{10} - \frac{3\sqrt{197}}{2}}$ .

The sum of the semi-major and semi-minor axes is therefore $a + b = \sqrt{\frac{211}{10} + \frac{3\sqrt{197}}{2}} + \sqrt{\frac{211}{10} - \frac{3\sqrt{197}}{2}} \approx \boxed{6.7082}$ .

Let $\mathbf{v_0} = \begin{bmatrix} 5 \\ 4 \\ 10 \end{bmatrix} \hspace{24pt} \mathbf{v_1} = \begin{bmatrix} 4 \\ 3 \\ -3 \end{bmatrix} \hspace{24pt} \mathbf{v_2} = \begin{bmatrix} 3 \\ 5 \\ 9 \end{bmatrix}$

Then, our ellipse is $\mathbf{p}(t) = \mathbf{v_0} + \mathbf{v_1} \cos t + \mathbf{v_2} \sin t \hspace{12pt} (1)$

Define the matrix $\mathbf{V} = [\mathbf{v_1} , \mathbf{v_2} ]$ and the vector $\mathbf{u}(t) = \begin{bmatrix} \cos t \\ \sin t \end{bmatrix}$

Then the equation of the ellipse is $\mathbf{p}(t) = \mathbf{v_0} + \mathbf{V u }(t) \hspace{12pt}(2)$

The parallel light rays along vector $\mathbf{d}$ together with the given ellipse generate an elliptical cylinder whose equation we want to find.

A point $\mathbf{q}$ on the surface of this cylinder is given by

$\mathbf{q}(t, s) = \mathbf{p}(t) + s \mathbf{d} = \mathbf{v_0} + \mathbf{V u}(t) + s \mathbf{d} \hspace{12pt} (3)$

Define the projection matrix $\mathbf{M} = \mathbf{I} - \mathbf{d d}^T$ , where $\mathbf{d}$ is assumed to have been normalized. Now pre-multiply both sides of equation $(3)$ by matrix $\mathbf{M}$ , you get,

$\mathbf{ M (q - v_0) }= \mathbf{ M V u } \hspace{12pt}(4)$

We want to solve for the vector $\mathbf{u}$ , so we'll multiply both sides by $\mathbf{(MV)}^T$ ,

$\mathbf{ V}^T \mathbf{M}^T \mathbf{M (q - v_0 )} = \mathbf{V}^T \mathbf{M}^T \mathbf{M V u } \hspace{12pt}(5)$

Noting that $\mathbf{M}^T = \mathbf{M}$ and $\mathbf{M}^2 = \mathbf{M}$ , this becomes,

$\mathbf{ V}^T \mathbf{M (q - v_0 )} = \mathbf{V}^T \mathbf{M V u } \hspace{12pt}(6)$

Pre-multiplying by the inverse of the $2 \times 2$ matrix $\mathbf{V}^T \mathbf{M V}$ , we get vector $\mathbf{u}$ explicitly.

$\mathbf{u} = ( \mathbf{V}^T \mathbf{M V})^{-1} \mathbf{ V}^T \mathbf{M (q - v_0 ) } = \mathbf{Q_1 (q - v0) } \hspace{12pt}(7)$

Now since $\mathbf{u}^T \mathbf{u} = 1$ , then the equation of the cylinder is given by

$\mathbf{(q - v_0)}^T \mathbf{Q (q - v_0) } = 1 \hspace{12pt}(8)$

where $\mathbf{Q} = \mathbf{Q_1}^T \mathbf{Q_1}$ .

Now that we have the equation of the cylinder, we can intersect it with any plane. If the plane is specified using the vector equation,

$\mathbf{r} = \mathbf{r_0 + U c} \hspace{12pt}(9)$

where $U$ is a $3 \times 2$ matrix whose columns are unit orthogonal vectors that together with $\mathbf{r_0}$ , span the plane under consideration. Vector $\mathbf{c} = [ c_1, c_2]^T$ is the coordinate vector of any point on the plane.

Substituting the expression for $\mathbf{r}$ into the equation of the cylinder, yields,

$\mathbf{(r_0 - v_0+ U c )}^T \mathbf{Q (r_0- v_0 + U c) } = 1 \hspace{12pt}(10)$

Let $\mathbf{w_0 = r_0 - v_0 }$ , then expanding equation $(10)$ , we get,

$\mathbf{c^T U^T Q U c + 2 c^T U^T Q w_0 + w_0^T Q w_0 } = 1 \hspace{12pt} (11)$

If we define the center point

$\mathbf{w_1} = - \mathbf{(U^T Q U)}^{-1} \mathbf{ U^T Q w_0 } \hspace{12pt} (12)$

then equation $(11)$ becomes,

$\mathbf{( c - w_1 )^T U^T Q U (c - w_1)} = \mathbf{1 - w_0^T Q w_0 + w_1 U^T Q U w_1 } \hspace{12pt} (13)$

Dividing by the right hand side, we obtain the standard equation of an ellipse in the plane,

$\mathbf{( c - w_1 )^T F (c - w_1)} = 1 \hspace{12pt} (14)$

where $\mathbf{F} = \dfrac{\mathbf{U^T Q U} }{\mathbf{1 - w_0^T Q w_0 + w_1 U^T Q U w_1 } } \hspace{12pt} (15)$

Naturally, matrix $\mathbf{F}$ is positive definite, and its two positive eigenvalues are the reciprocal of the squares of the semi-major and semi-minor axes lengths of the shadow ellipse on the plane. Performing all of the above calculations using a suitable application (for example MATLAB) or Wolframalpha.com , one finds the semi-major axis $a$ and semi-minor axis $b$ are given by

$a = 6.492572932$ and $b = 0.215631001$

And thus the answer is $a + b \approx \boxed{ 6.7082 }$