jmc

We have that $D=(3,0)$ and $B=(k-3,0).$ Hence, the $x$-coordinates of $A$ and $C$ are $\frac{k}{2}.$ The length of diagonal $BD$ is $6-k,$ so the $y$-coordinate of $A$ is $\frac{6-k}{2}.$ Hence, $$\frac{(k/2{)}^2}{9}+{\left(\frac{6-k}{2}\right)}^2=1.$$This simplifies to $5k^2-54k+144=0,$ which factors as $(k-6)(5k-24)=0.$ Hence, $k=\boxed{\frac{24}{5}}.$