This is a problem from Apostol's Calculus, which I have difficulty solving.
If $r_1$ and $r_2$ denote the distances from a point $(x,y)$ on an ellipse to its foci, show that the equation $r_1+r_2= \text{constant}$ (satisfied by these distances) implies the relation
$$\mathbf{T}\cdot \nabla(r_1+r_2)=0,$$
where $\mathbf{T}$ is the unit tangent to the curve. Interpret this result geometrically, thereby showing that the tangent makes equal angles with the lines joining $(x,y)$ to the foci.
I would greatly appreciate any solutions or suggestions.
Parametrize the ellipse. You can actually do this for any curve, so let's just call the parametrization $\gamma$ without specifying it anyhow. Let $x_0$ be a point on the ellipse and $t_0$ such that $x_0 = \gamma(t_0)$.
Now the function $t\to (r_1+r_2)(\gamma(t))$ is constant, since $\gamma(t)$ always lies on the ellipse and $r_1+r_2$ is constant there. So its derivative is zero. But using the chain rule its derivative is
$$\nabla(r_1+r_2)(\gamma(t))\cdot\gamma'(t)$$
Set $t=t_0$ and remember that $\gamma'(t)$ is in the direction of the tangent of the curve at the valuated point. This gives the equation.