[THOUGHT] Let me analyze this problem step by step.

The hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{a^2} = -1$, which can be rewritten as $\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1$.

This is a hyperbola with vertical transverse axis where $a^2 = a^2$ and $b^2 = a^2$, so $c^2 = a^2 + a^2 = 2a^2$, meaning $c = a\sqrt{2}$.

The foci are at $F = (0, a\sqrt{2})$ and $F' = (0, -a\sqrt{2})$.

First, let me verify that $(a, \sqrt{2}a)$ is on the hyperbola: $\frac{(\sqrt{2}a)^2}{a^2} - \frac{a^2}{a^2} = \frac{2a^2}{a^2} - 1 = 2 - 1 = 1$ ✓

Now I need to find the tangent line at $(a, \sqrt{2}a)$.

For the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1$, the tangent at point $(x_0, y_0)$ is: $\frac{y \cdot y_0}{a^2} - \frac{x \cdot x_0}{a^2} = 1$

At $(a, \sqrt{2}a)$: $\frac{y \cdot \sqrt{2}a}{a^2} - \frac{x \cdot a}{a^2} = 1$ $\frac{\sqrt{2}y}{a} - \frac{x}{a} = 1$ $\sqrt{2}y - x = a$

To find where this meets the y-axis (x = 0): $\sqrt{2}y = a$ $y = \frac{a}{\sqrt{2}} = \frac{a\sqrt{2}}{2}$

So $P = (0, \frac{a\sqrt{2}}{2})$.

Now I need to find $\overline{PF} \times \overline{PF'}$.

$PF = |y_P - y_F| = |\frac{a\sqrt{2}}{2} - a\sqrt{2}| = |-\frac{a\sqrt{2}}{2}| = \frac{a\sqrt{2}}{2}$

$PF' = |y_P - y_{F'}| = |\frac{a\sqrt{2}}{2} - (-a\sqrt{2})| = |\frac{a\sqrt{2}}{2} + a\sqrt{2}| = \frac{3a\sqrt{2}}{2}$

$PF \times PF' = \frac{a\sqrt{2}}{2} \times \frac{3a\sqrt{2}}{2} = \frac{3 \cdot 2 \cdot a^2}{4} = \frac{3a^2}{2} = 8$

So $a^2 = \frac{16}{3}$.

Let me verify this with Python. [/THOUGHT] [PYTHON] import sympy as sp

a = sp.Symbol('a', positive=True)

Hyperbola: y^2/a^2 - x^2/a^2 = 1

Foci at (0, c) and (0, -c) where c = sqrt(a^2 + a^2) = a*sqrt(2)

c = a * sp.sqrt(2) F = (0, c) F_prime = (0, -c)

Point on hyperbola: (a, sqrt(2)*a)

x0, y0 = a, sp.sqrt(2)*a

Verify point is on hyperbola

check = y02/a2 - x02/a2 print(f"Point on hyperbola check (should be 1): {sp.simplify(check)}")

Tangent line at (x0, y0): yy0/a^2 - xx0/a^2 = 1

At x = 0: y*y0/a^2 = 1 => y = a^2/y0

y_P = a**2 / y0 y_P = sp.simplify(y_P) print(f"P = (0, {y_P})")

Calculate PF and PF'

PF = sp.Abs(y_P - F[1]) PF_prime = sp.Abs(y_P - F_prime[1])

PF = sp.simplify(y_P - F[1]) # Need to determine sign PF_prime = sp.simplify(y_P - F_prime[1])

print(f"y_P = {y_P}") print(f"y_F = {F[1]}") print(f"y_F' = {F_prime[1]}")

y_P = asqrt(2)/2, y_F = asqrt(2), y_F' = -a*sqrt(2)

PF = |asqrt(2)/2 - asqrt(2)| = |-asqrt(2)/2| = asqrt(2)/2

PF' = |asqrt(2)/2 + asqrt(2)| = |3asqrt(2)/2| = 3asqrt(2)/2

PF_val = sp.Abs(asp.sqrt(2)/2 - asp.sqrt(2)) PF_prime_val = sp.Abs(asp.sqrt(2)/2 + asp.sqrt(2))

PF_simplified = sp.simplify(PF_val) PF_prime_simplified = sp.simplify(PF_prime_val)

print(f"PF = {PF_simplified}") print(f"PF' = {PF_prime_simplified}")

Product

product = PF_simplified * PF_prime_simplified product = sp.simplify(product) print(f"PF * PF' = {product}")

Solve product = 8

eq = sp.Eq(product, 8) solution = sp.solve(eq, a) print(f"Solutions for a: {solution}")

Find a^2

for sol in solution: if sol > 0: a_squared = sol**2 print(f"a = {sol}, a^2 = {sp.simplify(a_squared)}") [/PYTHON]

[THOUGHT] The Python calculation confirms my analysis. The hyperbola has foci at $(0, a\sqrt{2})$ and $(0, -a\sqrt{2})$. The tangent at $(a, \sqrt{2}a)$ meets the y-axis at $P = (0, \frac{a\sqrt{2}}{2})$.

The distances are:

• $PF = \frac{a\sqrt{2}}{2}$
• $PF' = \frac{3a\sqrt{2}}{2}$

The product $PF \times PF' = \frac{3a^2}{2} = 8$, which gives $a^2 = \frac{16}{3}$.

Since $a^2 = \frac{p}{q} = \frac{16}{3}$, we have $p = 16$ and $q = 3$.

Therefore, $p + q = 16 + 3 = 19$. [/THOUGHT]

\boxed{19}