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 $2.4$:
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 
[![SymDiff13not][1]][1]
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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 $1.29$:
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 
[![SymDiff13area3][2]][2]
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  [1]: https://i.sstatic.net/Sk1mH.jpg
  [2]: https://i.sstatic.net/sPAq6.jpg