Still Not the Deathly Hallows Symbol

Let the $x$ -axis be the base of the triangle. Then the horizontal ellipse will have equation $\frac{x^2}{a^2} + \frac{(y-b)^2}{b^2} \; = \; 1$ and this will be tangential to the slant lines $y \; =\; 2a \pm x\sqrt{3}$ A standard result, effectively derived by @Pi Han Goh , states that a line $y = mx + c$ will be tangential to the ellipse $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1$ provided that $a^2m^2 + b^2 = c^2$ . In this case, we deduce that $3a^2 + b^2 \; = \; (2a - b)^2$ and hence that $a = 4b$ , which makes the eccentricity of the ellipse equal to $e=\tfrac14\sqrt{15}$ , giving $\lfloor 10^6e\rfloor = \boxed{968245}$ .