A line of Neutrino is moving on the shape. When it moves through something like water, It makes a cone (Cherenkov radiation) which as you see, the line making the cone has angle $\phi$ with the line of light. It makes a circle with radius $a$ on the perpendicular plane of the line of light. now think it has an angle $i$ with the plane we want to project the shape onto. So now, It makes an ellipse (Large half axis $a$ and small half axis $b$). can someone explain how will the ellipse looks like. I somewhere read we will get $\frac b a = \cos{i}$ but don't know why. (The vertex of the cone is on the same point on both figures)
Well, the eccentricity of the ellipse would be: $$e=\frac{\sin i}{\cos \phi}$$ The major axis of the ellipse would be $BC=BD+DC$ (see figure above). We can find $BD$ by seeing that $\angle ABD = \frac {\pi}{2} - \phi + i$ and using the sine rule: $$\frac{\sin(\frac{\pi}{2}-\phi+i)}{d}=\frac{\sin\phi}{BD}$$ giving $$BD=\frac{d\sin\phi}{\sin(\frac{\pi}{2}-\phi+i)}$$ Similarly we see $\angle ACD=\frac {\pi}{2} - \phi - i$ giving: $$DC=\frac{d\sin\phi}{\sin(\frac{\pi}{2}-\phi-i)}$$ Adding together and simplifying a bit we find that: $$BC = d\sin\phi \left(\frac{1}{\cos(\phi - i)}+ \frac{1}{\cos(\phi + i)}\right)$$ So, the semi-major axis is $a=\frac{BC}{2}$. The semi-minor axis $b$ can be found with the formula $b=a\sqrt{1-e^2}$.