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

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 area of the ellipse is $A = \pi ab = \pi \cdot \sqrt{\frac{211}{10} + \frac{3\sqrt{197}}{2}} \cdot \sqrt{\frac{211}{10} + \frac{3\sqrt{197}}{2}} = \pi \cdot \sqrt{\frac{211^2}{10^2} - \frac{3^2 \cdot 197}{2^2}} = \frac{7}{5} \pi$ . Therefore, $p = 7$ , $q = 5$ , and $p + q = \boxed{12}$ .

Area of original ellipse is $A_1 = \pi a b = \pi \sqrt{4^2 + 3^2 +(-3)^2 } \sqrt{ 3^2 + 5^2 + 9^2} = \pi \sqrt{34} \sqrt{115}$

Using cross product, the normal to the plane of the ellipse is along $(4, 3, -3) \times (3, 5, 9) = (42, -45, 11)$

and the acute angle this normal makes with vector $\mathbf{d}$ is $\theta_1 = \cos^{-1} \dfrac{ 7 }{ \sqrt{34 ( 115 )} \sqrt{30} }$

hence the area of the cross section of the elliptical cylinder of the rays blocked by the original ellipse is

$A_{\text{cross section}} = A_1 \cos \theta_1 = \dfrac{7}{ \sqrt{30} } \pi$

finally the area of the shadow ellipse in the $xy$ plane is $A_\text{shadow} = \dfrac{A_\text{cross section} }{ \cos \theta_2 }$

where $\theta_2$ is the acute angle between $\mathbf{d}$ and vector $\mathbf{k}$ (the unit normal vector to the $xy$ plane). Its cosine is given by

$\cos \theta_2 = \dfrac{5}{ \sqrt{30}}$

Thus, $A_\text{shadow} = \dfrac{7}{5} \pi$

so that the answer is $7+ 5 = \boxed{12}$