Question 13

For ellipse $E$: $a^2=18, b^2=12 \Rightarrow e_E = \sqrt{1 - \frac{12}{18}} = \frac{1}{\sqrt{3}}$. Foci are $(\pm ae, 0) = (\pm\sqrt{18} \cdot \frac{1}{\sqrt{3}}, 0) = (\pm\sqrt{6}, 0)$. For hyperbola $H$: $e_H = \sqrt{3}$. Same foci $(\pm\sqrt{6}, 0) \Rightarrow a_H e_H = \sqrt{6} \Rightarrow a_H\sqrt{3} = \sqrt{6} \Rightarrow a_H = \sqrt{2}$. $b_H^2 = a_H^2(e_H^2 - 1) = 2(3-1) = 4$. Equation of $H$: $\frac{x^2}{2} - \frac{y^2}{4} = 1$. Intersection with $x^2 = \sqrt{5}y$: $\frac{\sqrt{5}y}{2} - \frac{y^2}{4} = 1 \Rightarrow y^2 - 2\sqrt{5}y + 4 = 0$. Roots are $y = \frac{2\sqrt{5} \pm 2}{2} = \sqrt{5} \pm 1$. Let $y_1 = \sqrt{5}+1$ and $y_2 = \sqrt{5}-1$. Both are positive. $x_1^2 = \sqrt{5}(\sqrt{5}+1) = 5+\sqrt{5}$ and $x_2^2 = \sqrt{5}(\sqrt{5}-1) = 5-\sqrt{5}$. In the first quadrant, $P = (\sqrt{5+\sqrt{5}}, \sqrt{5}+1)$ and $Q = (\sqrt{5-\sqrt{5}}, \sqrt{5}-1)$. $d^2 = (\sqrt{5+\sqrt{5}} - \sqrt{5-\sqrt{5}})^2 + ((\sqrt{5}+1) - (\sqrt{5}-1))^2$ $d^2 = (5+\sqrt{5} + 5-\sqrt{5} - 2\sqrt{25-5}) + 2^2 = 10 - 2\sqrt{20} + 4 = 14 - 4\sqrt{5}$. Here $a=14$ and $b=-4$. Thus $a - b = 14 - (-4) = 18$. Final Answer: 18.