The figure also shows the orthogonal projection $\tilde{C}$ of C onto a second plane $\tilde{P}$ through T, the angle between the planes being α. As you probably know [or exercise] $\tilde{C}$ is in fact an ellipse, the original circle C having been compressed in a direction perpendicular to T. Thus it is clear that the curvature $\tilde{κ}$ of $\tilde{C}$ at p is less than the original curvature κ.
How do I do the exercise i.e: prove that the circle gets sent to an ellipse under orthogonal projection? This is the first time the book introduces orthogonal projections ( I think?) and I am not sure what relevant properties of it can be used to prove the situation.
What I know so far:
$$ \tilde{\sigma}= \sigma \cos \alpha$$