Not an answer, just an illustration to accompany the question. $K$ is an isosceles triangle with base $2$ and altitude $3$ (and so area $3$). First, I mistakenly computed the ellipse $E$ of *any area* with the smallest area symmetric difference with $K$. It has area about $2.4$: <hr /> [![SymDiff13not][1]][1] <hr /> After Gerhard's comment, I recomputed constraining $E$ to have area $3$. Then its center is $(0,1)$ and its axes are roughly $0.74$ and $1.29$: <hr /> [![SymDiff13area3][2]][2] <hr /> [1]: https://i.sstatic.net/Sk1mH.jpg [2]: https://i.sstatic.net/sPAq6.jpg