[THOUGHT] We are given the hyperbola equation: $\frac{x^2}{a^2} - \frac{y^2}{a^2} = -1$. Multiply by $a^2$: $x^2 - y^2 = -a^2$, or equivalently $y^2 - x^2 = a^2$. This is a hyperbola centered at the origin with transverse axis along the y-axis (since positive term on y^2). Standard form: $\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1$. Indeed, dividing by $a^2$ gives $\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1$. So it's a rectangular hyperbola? Actually $a^2$ same for both axes, so asymptotes slopes ±1.

Foci: For hyperbola $\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1$, distance from center to foci along y-axis: $c = \sqrt{a^2 + a^2} = a\sqrt{2}$. So foci are at $(0, \pm a\sqrt{2})$. Let $F = (0, a\sqrt{2})$ and $F' = (0, -a\sqrt{2})$.

Point on hyperbola: $(a, \sqrt{2}a)$. Check: plug into equation: $a^2/a^2 - (\sqrt{2}a)^2/a^2 = 1 - 2 = -1$, yes satisfies.

Tangent line at that point: We can find derivative implicitly. Given $x^2 - y^2 = -a^2$. Differentiate: $2x - 2y y' = 0$ → $y' = x/y$. At $(a, \sqrt{2}a)$, slope $m = a/(\sqrt{2}a) = 1/\sqrt{2}$. So tangent line: $y - \sqrt{2}a = (1/\sqrt{2})(x - a)$. Find intersection with y-axis (x=0): $y = \sqrt{2}a + (1/\sqrt{2})(-a) = \sqrt{2}a - a/\sqrt{2} = \sqrt{2}a - (\sqrt{2}/2)a?$ Wait: $a/\sqrt{2} = a\sqrt{2}/2$. So $y = \sqrt{2}a - \frac{\sqrt{2}}{2}a = \frac{\sqrt{2}}{2}a(2 - 1) = \frac{\sqrt{2}}{2}a = a/\sqrt{2}$. So P = (0, a/√2).

Now distances PF and PF': F = (0, a√2), F' = (0, -a√2). PF = distance between (0, a/√2) and (0, a√2): vertical distance = |a√2 - a/√2| = a|√2 - 1/√2| = a| (2 - 1)/√2 | = a(1/√2) = a/√2. PF' = distance between (0, a/√2) and (0, -a√2): vertical distance = |a/√2 - (-a√2)| = a/√2 + a√2 = a(1/√2 + √2) = a(1/√2 + √2) = a( (1 + 2)/√2 ) = a(3/√2) = 3a/√2.

Thus PF × PF' = (a/√2) × (3a/√2) = 3a^2 / 2.

Given PF × PF' = 8, so 3a^2 / 2 = 8 → a^2 = 16/3 → a = 4/√3 (positive real). So a^2 = 16/3.

Then problem says: If $a^2 = \dfrac{p}{q}$, compute $p+q$. So p=16, q=3, p+q=19.

But wait: check if P is indeed (0, a/√2). Let's compute numerically with Python to verify distances and product.

Also need to ensure we didn't miss sign: PF = |a√2 - a/√2| = a(√2 - 1/√2) = a( (2-1)/√2 ) = a/√2. Yes positive. PF' = |a/√2 - (-a√2)| = a/√2 + a√2 = a(1/√2 + √2) = a( (1+2)/√2 ) = 3a/√2. So product = 3a^2/2.

Set equal to 8: a^2 = 16/3.

Let's verify with Python. [/THOUGHT] [PYTHON] import sympy as sp

define a as positive real

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

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

tangent slope: derivative implicitly: x^2 - y^2 = -a^2

differentiate: 2x - 2y*y' = 0 => y' = x/y

slope = a / (sp.sqrt(2)*a) # simplifies to 1/sqrt(2) slope_simplified = sp.simplify(slope) print("slope:", slope_simplified)

tangent line: y - sqrt(2)a = slope(x - a)

intersection with y-axis: x=0

y_intercept = sp.sqrt(2)*a + slope * (-a) y_intercept_simplified = sp.simplify(y_intercept) print("y-coordinate of P:", y_intercept_simplified)

foci: (0, asqrt(2)) and (0, -asqrt(2))

F = (0, asp.sqrt(2)) Fprime = (0, -asp.sqrt(2)) P = (0, y_intercept_simplified)

distances PF and PF'

PF = sp.sqrt((P[0] - F[0])**2 + (P[1] - F[1])**2) PFprime = sp.sqrt((P[0] - Fprime[0])**2 + (P[1] - Fprime[1])**2) PF_simplified = sp.simplify(PF) PFprime_simplified = sp.simplify(PFprime) print("PF:", PF_simplified) print("PF':", PFprime_simplified)

product

product = sp.simplify(PF_simplified * PFprime_simplified) print("PF * PF':", product)

set equal to 8 and solve for a^2

equation = sp.Eq(product, 8) solution = sp.solve(equation, a**2) print("a^2 =", solution)

solution should be 16/3

print("a^2 simplified:", [sp.nsimplify(sol) for sol in solution])

compute p+q if a^2 = p/q

a_sq = solution[0] p = sp.numer(a_sq) q = sp.denom(a_sq) print("p =", p, "q =", q) print("p+q =", p + q) [/PYTHON]

[THOUGHT] The Python output confirms that $a^2 = \frac{16}{3}$, so $p = 16$ and $q = 3$. Thus $p + q = 19$. [/THOUGHT] $\boxed{19}$