Eccentricity of a Hyperbola

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,

$$\boxed{18}$$