For the ellipse
$$\frac{x^2}{18}+\frac{y^2}{12}=1$$
we have
$$a^2=18,\qquad b^2=12$$
Hence,
$$c^2=a^2-b^2=6$$
and the eccentricity is
$$e=\frac{c}{a}=\frac{\sqrt6}{3\sqrt2}=\frac1{\sqrt3}$$
Therefore, the hyperbola has eccentricity
$$e_H=\frac1e=\sqrt3$$
and the same foci, so
$$c^2=6$$
For the hyperbola,
$$e_H=\frac{c}{a_H}=\sqrt3$$
Hence,
$$a_H^2=\frac{c^2}{e_H^2}=\frac6{3}=2$$
Also,
$$b_H^2=c^2-a_H^2=6-2=4$$
Therefore, the equation of the hyperbola is
$$\frac{x^2}{2}-\frac{y^2}{4}=1$$
The parabola is
$$\sqrt5,y=x^2$$
or
$$y=\frac{x^2}{\sqrt5}$$
Substituting into the hyperbola,
$$\frac{x^2}{2}-\frac1{4}\left(\frac{x^2}{\sqrt5}\right)^2=1$$
$$\frac{x^2}{2}-\frac{x^4}{20}=1$$
Multiplying by $$20$$,
$$10x^2-x^4=20$$
$$x^4-10x^2+20=0$$
Let
$$t=x^2$$
Then
$$t^2-10t+20=0$$
$$t=5\pm\sqrt5$$
Hence the two points are
$$P\left(\sqrt{5-\sqrt5},\frac{5-\sqrt5}{\sqrt5}\right)$$
and
$$Q\left(\sqrt{5+\sqrt5},\frac{5+\sqrt5}{\sqrt5}\right)$$
Now,
$$d^2=(x_2-x_1)^2+(y_2-y_1)^2$$
First,
$$y_2-y_1=\frac{(5+\sqrt5)-(5-\sqrt5)}{\sqrt5}=2$$
Also,
$$x_2^2+x_1^2=10$$
and
$$x_1x_2=\sqrt{(5-\sqrt5)(5+\sqrt5)}=\sqrt{20}=2\sqrt5$$
Therefore,
$$ (x_2-x_1)^2=x_2^2+x_1^2-2x_1x_2 $$
$$=10-4\sqrt5$$
Hence,
$$d^2=(10-4\sqrt5)+4$$
$$=14-4\sqrt5$$
Thus,
$$a=14,\qquad b=-4$$
and
$$a-b=14-(-4)$$
$$=18$$
Hence,
$$\boxed{18}$$
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