Value of Integer Expression for Distance Between Curves

The equation of the given ellipse $E$ is: $\frac{x^2}{18} + \frac{y^2}{12} = 1$

Here, $A^2 = 18$ and $B^2 = 12$. The eccentricity of ellipse $E$, denoted by $e_E$, is: $e_E = \sqrt{1 - \frac{B^2}{A^2}} = \sqrt{1 - \frac{12}{18}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$

The foci of $E$ are given by $(\pm A e_E, 0)$: $\pm A e_E = \pm \sqrt{18} \cdot \frac{1}{\sqrt{3}} = \pm \sqrt{6}$ Thus, the foci are $(\pm \sqrt{6}, 0)$.

Let the hyperbola $H$ be defined as: $\frac{x^2}{a_H^2} - \frac{y^2}{b_H^2} = 1$

The eccentricity of $H$, denoted by $e_H$, is the reciprocal of $e_E$: $e_H = \frac{1}{e_E} = \sqrt{3}$

Since $H$ shares the same foci as $E$, the focal distance $c = a_H e_H = \sqrt{6}$. Using $e_H = \sqrt{3}$: $a_H \sqrt{3} = \sqrt{6} \implies a_H = \sqrt{2} \implies a_H^2 = 2$

Now, using the eccentricity relation for a hyperbola $b_H^2 = a_H^2 (e_H^2 - 1)$: $b_H^2 = 2 ((\sqrt{3})^2 - 1) = 2(3 - 1) = 4$

Thus, the equation of the hyperbola $H$ is: $\frac{x^2}{2} - \frac{y^2}{4} = 1$

To find the points of intersection of $H$ with the parabola $x^2 = \sqrt{5}y$, substitute $x^2 = \sqrt{5}y$ into the equation of $H$: $\frac{\sqrt{5}y}{2} - \frac{y^2}{4} = 1$

Multiplying by $4$ gives the quadratic equation: $y^2 - 2\sqrt{5}y + 4 = 0$

Solving for $y$ using the quadratic formula: $y = \frac{2\sqrt{5} \pm \sqrt{(-2\sqrt{5})^2 - 4(1)(4)}}{2} = \frac{2\sqrt{5} \pm \sqrt{20 - 16}}{2} = \sqrt{5} \pm 1$

Thus, the two $y$-coordinates are: $y_1 = \sqrt{5} + 1 \quad \text{and} \quad y_2 = \sqrt{5} - 1$

Since $x^2 = \sqrt{5}y$, the corresponding $x$-coordinates for the first quadrant ($x > 0, y > 0$) are: $x_1 = \sqrt{\sqrt{5}(\sqrt{5} + 1)} = \sqrt{5 + \sqrt{5}}$ $x_2 = \sqrt{\sqrt{5}(\sqrt{5} - 1)} = \sqrt{5 - \sqrt{5}}$

Therefore, the points of intersection in the first quadrant are: $P = \left(\sqrt{5 + \sqrt{5}}, \sqrt{5} + 1\right) \quad \text{and} \quad Q = \left(\sqrt{5 - \sqrt{5}}, \sqrt{5} - 1\right)$

The square of the distance $d$ between $P$ and $Q$ is given by: $d^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2$

Calculating the terms separately: $(y_1 - y_2)^2 = ((\sqrt{5} + 1) - (\sqrt{5} - 1))^2 = 2^2 = 4$

$(x_1 - x_2)^2 = x_1^2 + x_2^2 - 2x_1 x_2$ $x_1^2 + x_2^2 = (5 + \sqrt{5}) + (5 - \sqrt{5}) = 10$ $x_1 x_2 = \sqrt{(5 + \sqrt{5})(5 - \sqrt{5})} = \sqrt{25 - 5} = \sqrt{20} = 2\sqrt{5}$

Substituting these values back into $(x_1 - x_2)^2$: $(x_1 - x_2)^2 = 10 - 4\sqrt{5}$

Therefore: $d^2 = (10 - 4\sqrt{5}) + 4 = 14 - 4\sqrt{5}$

Comparing this with $d^2 = a + b\sqrt{5}$, we identify the integers $a$ and $b$ as: $a = 14, \quad b = -4$

Thus, the value of $a - b$ is: $a - b = 14 - (-4) = 18$