A simple Steiner's ellipse

Consider an equilateral triangle with two of its vertices at $A(-a, 0)$ and $B(a, 0)$ , its top vertex at $C$ , and its center at $D$ . By the properties of an equilateral triangle, the center will be $\frac{1}{3}$ of the way up the height.

Now stretch the equilateral triangle and its circumcircle so that $C$ moves to $C'(0, h)$ . The stretched circumcircle is now the Steiner ellipse (the ellipse of minimum area passing through the vertices of $\triangle ABC$ ), and its new center will still be $\frac{1}{3}$ of the way up its height at $D'(0, \frac{1}{3}h)$ .

Therefore, when $h = 12$ , the $y$ -coordinate of the ellipse's center is $\frac{1}{3}h = \frac{1}{3}\cdot 12 = \boxed{4}$ .