If the tangent at any point of the ellipse make an angle $\alpha$ with the major axis and an angle $\beta$ with the focal radius of the point of contact, then show that the eccentricity of the ellipse is given by $$e=\dfrac{\cos\beta}{\cos\alpha}$$
How is this derived? Please explain with a proper diagram.
Let $F$ and $G$ be the foci of the ellipse. The bisector $PB$ of $\angle FPG$ is perpendicular to tangent $PQ$. Hence: $$ \begin{align} e&={FB+GB\over FP+GP}\quad\text{(definition)}\\ &={FB\over FP}\quad\text{(angle bisector theorem)}\\ &={\sin(\pi/2-\beta)\over\sin(\pi/2-\alpha)}\quad\text{(sine rule)}\\ &={\cos\beta\over\cos\alpha}. \end{align} $$
