= Solution
Let $\mathbf r$ be the orbiting <point mass>'s <position> relative to the attracting body, and let its <velocity> be $\mathbf v=\dot{\mathbf r}$. Its <angular momentum> is
$$
\mathbf L=m\mathbf r\times\dot{\mathbf r}.
$$
<Conservation of angular momentum> keeps both the magnitude and direction of $\mathbf L$ fixed. Since $\mathbf L$ is perpendicular to $\mathbf r$, a nonradial orbit lies in a fixed <plane>. During a short interval $dt$, the radius vector sweeps a thin <triangle> of <area>
$$
dA=\frac12|\mathbf r\times d\mathbf r|=\frac12|\mathbf r\times\dot{\mathbf r}|\,dt.
$$
Thus the <areal velocity> is constant:
$$
\boxed{\frac{dA}{dt}=\frac{L}{2m},\qquad \Delta A=\frac{L}{2m}\,\Delta t.}
$$
Equal <time> intervals give equal swept <areas>: this is <Kepler's second law>. In <plane polar coordinates>, the same statement is $L=mr^2\dot\phi$ and $\dot A=r^2\dot\phi/2$. The attracting body is at the relevant focus of the <ellipse>, not its geometric centre. The area argument needs conserved <angular momentum>, not specifically an <ellipse>; a <central force> provides this conservation because its <torque> $\mathbf r\times\mathbf F$ vanishes.
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