Connecting tangents of convex curves: at some point orthogonal?

Question

Let $a(t)$ and $b(t)$ be two smooth, nested convex curves in the plane, $t\in[0,1]$:
     
Suppose the parametrization of $a()$ and $b()$ is such that $\dot{a}(t)$ is parallel to $\dot{b}(t)$: at time $t$, the tangents at $a(t)$ and $b(t)$ are parallel. Let $ab(t) = b(t)-a(t)$ be the vector from $a(t)$ to $b(t)$, and let $\theta(t)$ be the counterclockwise angle from $ab(t)$ to $\dot{a}(t)$. I would like to claim that

$\theta(t) = \pi/2$ at least twice within $t\in[0,1]$.

My proof of this is inelegant, relying on the length $|ab(t)|$, essentially showing that if, e.g, $\theta(t) < \pi/2$ for all $t$, then $|ab(1)| >|ab(0)|$ (contradicting $|ab(1)| = |ab(0)|$). But I feel there might be a clever way to achieve this via the mean-value theorem that I am not seeing. Also, perhaps "twice" can be replaced by "four times," and perhaps "nested" need not be assumed. So I am seeking a cleaner proof that may yield further insights. Thanks for contributing ideas or pointing me in the right direction!

Answer 1

Take points $x, y\in b$ which minimize (correspondingly maximize) the function $\mathop{\rm dist}_a$. Let $\bar x$ and $\bar y\in a$ be the closest points to $x$ and $y$ correspondingly.

Clearly $\bar x=a(t)$, $x=b(t)$ and $\bar y=a(\tau)$, $y=b(\tau)$ for some $t$ and $\tau$ and $\theta(t)=\theta(\tau)=\tfrac\pi2$.

If $a$ and $b$ are non-concentric circles then you get only these two values.