From hyperbola conjugate semi-diameters to semi-axes lengths

$\mathbf{C}$ only shifts the graph left and right, so we can let $\mathbf{C} = (0, 0)$ . Then with $\mathbf{f_1} = (3, 1)$ and $\mathbf{f_2} = (0, 2)$ , we have parametric equations:

$x = 3 \sec t$

$y = \sec t + 2 \tan t$

With a little bit of rearranging, we can eliminate $t$ and obtain:

$x^2 + 2xy - 3y^2 - 12 = 0$

a conic where $A = 1$ , $B = 2$ , $C = -3$ , and $F = -12$ , and is rotated $\theta$ , where $\cot 2\theta = \frac{A - C}{B} = \frac{1 - -3}{2} = 2$ . Using some trig identities, this means that:

$\sin \theta = \sqrt{\frac{1}{10}(5 - 2\sqrt{5})}$

$\cos \theta = \sqrt{\frac{1}{10}(5 + 2\sqrt{5})}$

and applying the tranformation rotation :

$A' = A \cos^2 \theta + B \sin \theta \cos \theta + C \sin^2 \theta = \sqrt{5} - 1$

$C' = A \sin^2 \theta - B \sin \theta \cos \theta + C \cos^2 \theta = -(\sqrt{5} + 1)$

$F' = F = -12$

so that a new graph with the same scale is:

$(\sqrt{5} - 1)x'^2 - (\sqrt{5} + 1)y'^2 - 12 = 0$

which rearranges to:

$\frac{x'^2}{\frac{12}{\sqrt{5} - 1}} - \frac{y'^2}{\frac{12}{\sqrt{5} + 1}} = 1$

so that

$a = \sqrt{\frac{12}{\sqrt{5} - 1}}$

$b = \sqrt{\frac{12}{\sqrt{5} + 1}}$

and $a + b = \sqrt{\frac{12}{\sqrt{5} - 1}} + \sqrt{\frac{12}{\sqrt{5} + 1}} = \sqrt{6(2 + \sqrt{5})} \approx \boxed{5.041}$ .