The hyperbola equation is (\frac{x^2}{a^2} - \frac{y^2}{a^2} = -1), which simplifies to (\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1). This is a hyperbola opening vertically with (a^2 = a^2) and (b^2 = a^2), so (c^2 = a^2 + b^2 = 2a^2), giving foci at ((0, a\sqrt{2})) and ((0, -a\sqrt{2})).

The point (P(a, \sqrt{2}a)) lies on the hyperbola. The tangent at this point is found using the formula for the tangent to (\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1) at ((x_1, y_1)): [ \frac{y y_1}{a^2} - \frac{x x_1}{b^2} = 1. ] Substituting (b^2 = a^2) and ((x_1, y_1) = (a, \sqrt{2}a)): [ \frac{y \cdot \sqrt{2}a}{a^2} - \frac{x \cdot a}{a^2} = 1 \implies \sqrt{2} y - \frac{x}{a} = 1. ] At the y-axis ((x = 0)): [ \sqrt{2} y = 1 \implies y = \frac{\sqrt{2}}{2}. ] So the tangent meets the y-axis at (Q(0, \frac{\sqrt{2}}{2})).

The distances from (Q) to the foci (F(0, a\sqrt{2})) and (F'(0, -a\sqrt{2})) are: [ \overline{QF} = \left| \frac{\sqrt{2}}{2} - a\sqrt{2} \right| = \sqrt{2} \left| \frac{1}{2} - a \right|, \quad \overline{QF'} = \left| \frac{\sqrt{2}}{2} + a\sqrt{2} \right| = \sqrt{2} \left( \frac{1}{2} + a \right). ] The product is: [ \overline{QF} \times \overline{QF'} = \left( \sqrt{2} \left| \frac{1}{2} - a \right| \right) \times \left( \sqrt{2} \left( \frac{1}{2} + a \right) \right) = 2 \left| \frac{1}{2} - a \right| \left( \frac{1}{2} + a \right). ] Simplifying the expression inside: [ \left| \frac{1}{2} - a \right| \left( \frac{1}{2} + a \right) = \left| \left( \frac{1}{2} - a \right) \left( \frac{1}{2} + a \right) \right| = \left| \frac{1}{4} - a^2 \right|. ] Thus: [ 2 \left| \frac{1}{4} - a^2 \right| = 8 \implies \left| a^2 - \frac{1}{4} \right| = 4. ] Solving: [ a^2 - \frac{1}{4} = 4 \quad \text{or} \quad a^2 - \frac{1}{4} = -4. ] The solutions are (a^2 = \frac{17}{4}) or (a^2 = -\frac{15}{4}). Since (a^2 > 0), only (a^2 = \frac{17}{4}) is valid. Thus, (p = 17), (q = 4), and (p + q = 21).

\boxed{21}

[THOUGHT] We will use Python to verify the product of the distances from the point Q to the foci for the value of a^2 = 17/4, to ensure it equals 8 as given in the problem.

[PYTHON]

Define a^2 and the y-coordinate of Q

a_sq = 17/4 y_q = (2**0.5) / 2

Calculate the y-coordinates of the foci

f1_y = a_sq0.5 * (20.5) f2_y = -a_sq0.5 * (20.5)

Compute the distances from Q to the foci

dist1 = abs(y_q - f1_y) dist2 = abs(y_q - f2_y)

Calculate the product of the distances

product = dist1 * dist2

Print the result

print(product) [/PYTHON]

\boxed{21}