No Einstein here! Only an innocent conic

As we can see here , the ellipse with implicit equation $A{{x}^{2}}+Bxy+C{{y}^{2}}+Dx+Ey+F=0$ is an isometric transformation (translation and rotation) of the ellipse with canonical equation $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ .

For the rotation angle, $\Theta$ , it holds the relation $\Theta =\arctan \left( \frac{1}{B}\left( C-A-\sqrt{{{\left( A-C \right)}^{2}}+{{B}^{2}}} \right) \right).$ Hence, the slope of the major axis is $m=\tan \Theta =\dfrac{1}{B}\left( C-A-\sqrt{{{\left( A-C \right)}^{2}}+{{B}^{2}}} \right)$ .
Using $A=14,\text{ }B=-4,\text{ }C=11,\text{ }D=-44,\text{ }E=-58,\text{ }F=71$ , we find $m=2$ . $$

$\ \ \$

Furthermore, $a=\dfrac{-\sqrt{2\left( A{{E}^{2}}+C{{D}^{2}}-BDE+\left( {{B}^{2}}-4AC \right)F \right)\left( \left( A+C \right)+\sqrt{{{\left( A-C \right)}^{2}}+{{B}^{2}}} \right)}}{{{B}^{2}}-4AC}=\sqrt{6}$

Thus, the sum of distances of any variable point on the ellipse from its focii is $c=2a=2\sqrt{6}$ .

For the answer, $m{{c}^{2}}=2\times {{\left( 2\sqrt{6} \right)}^{2}}=\boxed{48}$ .