Consider the ellipse E given by (x^2)/(18) + (y^2)/(12) = 1 . Let H...

For the ellipse $E: \dfrac{x^2}{18} + \dfrac{y^2}{12} = 1$, we have $a_E^2 = 18$ and $b_E^2 = 12$.

The eccentricity $e_E$ is given by:
$e_E = \sqrt{1 - \dfrac{b_E^2}{a_E^2}} = \sqrt{1 - \dfrac{12}{18}} = \dfrac{1}{\sqrt{3}}$

The foci of the ellipse $E$ are $(\pm a_E e_E, 0) = \left(\pm \sqrt{18} \times \dfrac{1}{\sqrt{3}}, 0\right) = (\pm \sqrt{6}, 0)$.

For the hyperbola $H$, the eccentricity $e_H$ is the reciprocal of $e_E$, so $e_H = \sqrt{3}$.
Since $H$ has the same foci as $E$, its foci are $(\pm \sqrt{6}, 0)$.
Let the equation of $H$ be $\dfrac{x^2}{A^2} - \dfrac{y^2}{B^2} = 1$.
The foci are $(\pm A e_H, 0)$, which gives $A \sqrt{3} = \sqrt{6} \Rightarrow A = \sqrt{2} \Rightarrow A^2 = 2$.
Also, $B^2 = A^2(e_H^2 - 1) = 2(3 - 1) = 4$.
Thus, the equation of the hyperbola $H$ is $\dfrac{x^2}{2} - \dfrac{y^2}{4} = 1$.

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

Multiplying by $4$, we get:
$2\sqrt{5}y - y^2 = 4 \Rightarrow y^2 - 2\sqrt{5}y + 4 = 0$

Solving for $y$ using the quadratic formula:
$y = \dfrac{2\sqrt{5} \pm \sqrt{20 - 16}}{2} = \sqrt{5} \pm 1$

Let $y_1 = \sqrt{5} - 1$ and $y_2 = \sqrt{5} + 1$. Both are positive.
The corresponding $x^2$ values are:
$x_1^2 = \sqrt{5}(\sqrt{5} - 1) = 5 - \sqrt{5}$
$x_2^2 = \sqrt{5}(\sqrt{5} + 1) = 5 + \sqrt{5}$

Since the points $P$ and $Q$ lie in the first quadrant, $x > 0$. Thus, $x_1 = \sqrt{5 - \sqrt{5}}$ and $x_2 = \sqrt{5 + \sqrt{5}}$.

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

First, we evaluate $(x_2 - x_1)^2$:
$(x_2 - x_1)^2 = x_2^2 + x_1^2 - 2x_1x_2$
$x_2^2 + x_1^2 = (5 + \sqrt{5}) + (5 - \sqrt{5}) = 10$
$x_1x_2 = \sqrt{(5 - \sqrt{5})(5 + \sqrt{5})} = \sqrt{25 - 5} = \sqrt{20} = 2\sqrt{5}$
So, $(x_2 - x_1)^2 = 10 - 4\sqrt{5}$.

Next, we evaluate $(y_2 - y_1)^2$:
$(y_2 - y_1)^2 = ((\sqrt{5} + 1) - (\sqrt{5} - 1))^2 = 2^2 = 4$

Adding these together gives $d^2$:
$d^2 = 10 - 4\sqrt{5} + 4 = 14 - 4\sqrt{5}$

Comparing this with $d^2 = a + b\sqrt{5}$, we get $a = 14$ and $b = -4$.

Therefore, $a - b = 14 - (-4) = 18$.

Answer: $18$