Circular Cross Sections

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

[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]

Circular Conic Section

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

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

Using the relation

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