Suppose $u$ and $v$ are continuous on $[a,b]$ and differentiable on $(a,b)$, and assume that for all $t\in(a,b)$, at least one of $u'(t)$ and $v'(t)$ is nonzero. Let $C$ be the curve given by $(u(t),v(t))$ for $t \in [a,b]$. Let $A = (u(a),v(a))$ and $B = (u(b),v(b))$ be the endpoints of the curve, and assume $A\ne B$. Show that there is some point $c \in (a,b)$ such that the tangent line to $C$ at $(u(c),v(c))$ is parallel to $\overline{AB}$.
I had to answer this last week, and I was struggling. I know that by the Mean Value Theorem we have that $\exists c : v'(c) = \frac{v(b) - v(a)}{b - a}$ and $\exists d : u'(d) = \frac{u(b) - u(a)}{b - a}$, which gives us $\frac{v(b) - v(a)}{u(b) - u(a)} = \frac{v'(c)}{u'(d)}$.
A friend gave me a hint that "Let $h(t) = v(t) - v(a) - \frac{v(b) - v(a)}{u(b) - u(a)}(u(t) - u(a))$. Remembering that $v(a)$, $\frac{v(b) - v(a)}{u(b) - u(a)}$, and $u(a)$ are constants, what is $h'(t)$?"
This is no longer for credit, I just want help understanding what I'm missing.
Hint: Try $$ h(t)=(u(b) - u(a))v(t) - (v(b) - v(a))u(t)$$ and then use Rolle's MVT.