Question: 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 $\overline{AB}$.
My attempt
We know the endpoints are: $$A = (u(a),v(a))$$ $$B = (u(b),v(b))$$ so the gradient of line $AB$ is: $$\frac{v(a) - v(b)}{u(a) - u(b)}$$
Now the gradient of a point $C$ at point $(u(c),v(c))$ is: $$\frac{v'(c)}{u'(c)}$$
Now we know Mean Value Theorem as: $$f'(c) = \frac{f(a) - f(b)}{a-b}$$
Now I don't know how to proceed. I first thought of using MVT for $v(t)$ and $u(t)$ and dividing but the two constant terms need not be equal. So I'm not sure which method to use.
Any help would be much appreciated!
Let $$f(x) = u(x)\Big(v(b)-v(a)\Big)-v(x)\Big(u(b)-u(a)\Big)$$ Then \begin{align} f(b) &= v(b)u(a)-u(b)v(a)\\ f(a)&=v(b)u(a)-u(b)v(a)\\ &=f(b) \end{align} Therefore, all assumptions of Rolle's theorem are satisfied by f(x) (continuity and differentiablity are obvious). Hence, there is a point $c$ in $[a,b]$ such that $f'(c)=0$. Hence, \begin{gather} u'(c)\Big(v(b)-v(a)\Big)-v'(c)\Big(u(b)-u(a)\Big)=0\\ \implies \dfrac{v(b)-v(a)}{u(b)-u(a)} = \dfrac{v'(c)}{u'(c)} \end{gather} Hence proved.