Suppose that an ellipse is rolling along a line. If we follow the path of one of the foci of the ellipse as it rolls, then this path formes a curve - namely an undulary. Now consider the following diagram
The points $F$ and $F'$ are the forci of the ellipse, which is rolling along the line $KT$. The line $FT$ is the tangent to the curve traced out by $F$. Now the following folds:
$FT$ is is perpendicular to $FK$, so the normal to the locus of $F$ passes through $K$.
Why is this always the case? I read this in 3 several books, but every time without a proof. Is it that obvious? Can anyone explain it or give a proof/source?
Best regards!
Let $F(t)$ be the curve described by point $F$ and $K(t)$ the curve described by point $K$. When $K$ is on the line (ground) its velocity is zero by assumption (rolling on the line): $$ K'(t) = 0. $$ But we know that $|F(t)-K(t)|$ is constant (the ellipse is rigid) so: $$ 0 = \frac{d}{dt} (F(t)-K(t))^2 = (F(t)-K(t)) \cdot (F'(t) - K'(t)) = (F(t) -K(t)) \cdot F'(t) $$ which means that the velocity of $F$ (which is along the tangent line to the curve $F(t)$) is orthogonal to the segment $FK$.