Inellipse of a triangle

If $B$ is $(0, 0)$ and $C$ is $(9, 0)$ (as pictured), and $A$ is $(A_x, A_y)$ , then by the distance equation:

$AB = \sqrt{(A_x - 0)^2 + (A_y - 0)^2} = 7$

$AC = \sqrt{(A_x - 9)^2 + (A_y - 0)^2} = 8$

which solves to $A_x = \frac{11}{3}$ and $A_y = \frac{8\sqrt{5}}{3}$ .

Since $BE = 0.3AB$ and $B$ is at $(0, 0)$ and $A$ is at $(\frac{11}{3}, \frac{8\sqrt{5}}{3})$ , the coordinates of $E$ are $(0.3 \cdot \frac{11}{3}, 0.3 \cdot \frac{8\sqrt{5}}{3}) = (\frac{11}{10}, \frac{4\sqrt{5}}{5})$ .

Since $BD = 0.3BC$ and $B$ is at $(0, 0)$ and $C$ is at $(9, 0)$ , the coordinates of $D$ are $(0.3 \cdot 9, 0.3 \cdot 0) = (\frac{27}{10}, 0)$ .

Using the equations from the link provided in the question, the points of contact are $(u_1, u_2) = (\frac{27}{10}, 0)$ and $(v_1, v_2) = (\frac{11}{10}, \frac{4\sqrt{5}}{5})$ , and $s = t = 0.3$ . That means:

$a = \frac{1}{s - 1} = \frac{1}{0.3 - 1} = -\frac{10}{7}$

$b = \frac{1}{t - 1} = \frac{1}{0.3 - 1} = -\frac{10}{7}$

the center of the inellipse is:

$G(x_0, y_0) = M = \frac{ab}{ab - 1}(\frac{u_1 + v_1}{2}, \frac{u_2 + v_2}{2}) = (\frac{190}{51}, \frac{40\sqrt{5}}{51})$

and the conjugate half-diameters of the ellipse are:

$\overrightarrow{f_1} = \frac{1}{2} \frac{\sqrt{ab}}{ab - 1}(u_1 + v_1, u_2 + v_2) = (\frac{133}{51}, \frac{28\sqrt{5}}{51})$

$\overrightarrow{f_2} = \frac{1}{2} \sqrt{\frac{ab}{ab - 1}}(u_1 - v_1, u_2 - v_2) = (\frac{8\sqrt{51}}{51}, -\frac{4\sqrt{255}}{51})$

Using the equation for the semi-axes lengths derived from this question :

$a = \sqrt{\frac{1}{2}(f_{1x}^2 + f_{1y}^2 + f_{2x}^2 + f_{2y}^2) + \frac{1}{2}\sqrt{((f_{1x} - f_{2y})^2 + (f_{1y} + f_{2x})^2)((f_{1x} + f_{2y})^2 + (f_{1y} - f_{2x})^2)}} = \sqrt{\frac{3217 + \sqrt{3151969}}{578}}$

$b = \sqrt{\frac{1}{2}(f_{1x}^2 + f_{1y}^2 + f_{2x}^2 + f_{2y}^2) - \frac{1}{2}\sqrt{((f_{1x} - f_{2y})^2 + (f_{1y} + f_{2x})^2)((f_{1x} + f_{2y})^2 + (f_{1y} - f_{2x})^2)}} = \sqrt{\frac{3217 - \sqrt{3151969}}{578}}$

Therefore, $x_0 + y_0 + a + b = \frac{190}{51} + \frac{40\sqrt{5}}{51} + \sqrt{\frac{3217 + \sqrt{3151969}}{578}} + \sqrt{\frac{3217 - \sqrt{3151969}}{578}} = \frac{190 +40\sqrt{5} + 3\sqrt{3217 + 168\sqrt{255}}}{51} \approx \boxed{9.997}$ .