The vertices of a hyperbola H are $$(\pm 6, 0)$$ and its eccentricity is $$\frac{\sqrt{5}}{2}$$. Let N be the normal to H at a point in the first quadrant and parallel to the line $$\sqrt{2}x + y = 2\sqrt{2}$$. If $$d$$ is the length of the line segment of N between H and the y-axis then $$d^2$$ is equal to _____.

Given: Hyperbola $$H$$ with vertices $$(\pm 6, 0)$$ and eccentricity $$e = \frac{\sqrt{5}}{2}$$.

So $$a = 6$$, $$c = ae = 3\sqrt{5}$$, $$b^2 = c^2 - a^2 = 45 - 36 = 9$$.

Hyperbola: $$\frac{x^2}{36} - \frac{y^2}{9} = 1$$.

Find the point on $$H$$ in the first quadrant where the normal has slope $$-\sqrt{2}$$ (parallel to $$\sqrt{2}x + y = 2\sqrt{2}$$).

For the hyperbola, $$y' = \frac{9x}{36y} = \frac{x}{4y}$$. Normal slope $$= -\frac{4y}{x} = -\sqrt{2}$$.

$$y = \frac{x\sqrt{2}}{4}$$

Substituting into the hyperbola equation:

$$\frac{x^2}{36} - \frac{x^2 \cdot 2}{16 \cdot 9} = 1 \implies \frac{x^2}{36} - \frac{x^2}{72} = 1 \implies \frac{x^2}{72} = 1 \implies x^2 = 72$$

$$x = 6\sqrt{2}$$, $$y = \frac{6\sqrt{2} \cdot \sqrt{2}}{4} = 3$$.

Normal at $$(6\sqrt{2}, 3)$$ with slope $$-\sqrt{2}$$:

$$y - 3 = -\sqrt{2}(x - 6\sqrt{2}) \implies y = -\sqrt{2}x + 12 + 3 = -\sqrt{2}x + 15$$

The normal meets the $$y$$-axis at $$(0, 15)$$. The line segment $$N$$ runs from $$(6\sqrt{2}, 3)$$ to $$(0, 15)$$.

$$d^2 = (6\sqrt{2})^2 + (15 - 3)^2 = 72 + 144 = \boxed{216}$$