Circular Cross Sections

[math]\underline{\textbf{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]

Circular Conic Section

If the conic is an circle, e=0. This implies

[math]e=\frac{\sin (\beta)}{\sin (\alpha)}=\frac{\sin (25^{\circ})}{\sin (90-\theta)}=0[/math]

Using the relation

[math]sin(90^{\circ}-\theta)=cos(\theta)[/math]

[math]\frac{sin (25^{\circ})}{0}=cos( \theta) =\infty[/math]

The sector angle will never be perpendicular to the plane of the light cone, so this is not a physical possibility.

---

[math]\underline{\textbf{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]