Consider the ellipse E given by $\frac{x^{2}}{18} + \frac{y^{2}}{12} = 1$. Let H be the hyperbola whose eccentricity is the reciprocal of the eccentricity of E and whose foci are the same as that of E. Let P and Q be the points of intersection of H and the parabola $\sqrt{5}y = x^{2}$ in the first quadrant. Let d be the distance between P and Q. If a and b are the integers such that $d^{2} = a + b\sqrt{5}$, then the value of $a - b$ is ______.

Consider the ellipse E given by x²/18 + y²/12 = 1. Let H be the hyperbola whose eccentricity is the reciprocal of the eccentricity of E and whose foci are the same as that of E. Let P and Q be the points of intersection of H and the parabola √5y = x² in the first quadrant. Let d be the distance between P and Q. If a and b are the integers such that d² = a + b√5, then the value of a - b is ______. $E: \dfrac{x^2}{18} + \dfrac{y^2}{12} = 1$ $e_H = \dfrac{1}{e_E}; \quad H \text{ and } E \text{ share the same foci}$ $H \cap \sqrt{5}\,y = x^2 \Rightarrow P, Q \ \text{(first quadrant)}$ $d = PQ; \quad d^2 = a + b\sqrt{5} \Rightarrow a - b = ?$ $E: a_E^2 = 18,\ b_E^2 = 12$ $1.\ e_E, \text{ foci of } E$ $2.\ H: e_H = 1/e_E,\ \text{same } c \Rightarrow A, B$ $3.\ H \cap (x^2 = \sqrt5 y) \Rightarrow P, Q \Rightarrow d^2$ $e_E^2 = 1 - \dfrac{b_E^2}{a_E^2} = 1 - \dfrac{12}{18} = \dfrac{1}{3} \Rightarrow e_E = \dfrac{1}{\sqrt3}$ $c^2 = a_E^2 - b_E^2 = 6 \Rightarrow c = \sqrt6,\ \text{foci } (\pm\sqrt6, 0)$ $e_H = \sqrt3 = \dfrac{c}{A} = \dfrac{\sqrt6}{A} \Rightarrow A = \sqrt2,\ A^2 = 2$ $B^2 = c^2 - A^2 = 6 - 2 = 4 \Rightarrow H: \dfrac{x^2}{2} - \dfrac{y^2}{4} = 1$ $\sqrt5\,y = x^2 \Rightarrow x^2 = \sqrt5\,y$ $\dfrac{x^2}{2} - \dfrac{y^2}{4} = 1,\quad x^2 = \sqrt5\,y$ $\dfrac{\sqrt5\,y}{2} - \dfrac{y^2}{4} = 1 \Rightarrow y^2 - 2\sqrt5\,y + 4 = 0$ $y = \dfrac{2\sqrt5 \pm \sqrt{20-16}}{2} = \dfrac{2\sqrt5 \pm 2}{2} = \sqrt5 \pm 1$ $y = \sqrt5+1 \Rightarrow x^2 = \sqrt5(\sqrt5+1) = 5+\sqrt5$ $y = \sqrt5-1 \Rightarrow x^2 = \sqrt5(\sqrt5-1) = 5-\sqrt5$ $d^2 = (x_1-x_2)^2 + (y_1-y_2)^2$ $= \left(\sqrt{5+\sqrt5} - \sqrt{5-\sqrt5}\right)^2 + 2^2$ $(x_1-x_2)^2 = (5+\sqrt5)+(5-\sqrt5) - 2\sqrt{(5+\sqrt5)(5-\sqrt5)}$ $= 10 - 2\sqrt{20} = 10 - 4\sqrt5$ $d^2 = 10 - 4\sqrt5 + 4 = 14 - 4\sqrt5$ $d^2 = 14 + (-4)\sqrt5 \Rightarrow a = 14,\ b = -4$ $\boxed{a - b = 14 - (-4) = 18}$