[THOUGHT] I need to find the value of $a$ for the given hyperbola. The hyperbola is $\dfrac{x^2}{a^2} - \dfrac{y^2}{a^2} = -1$, which can be rewritten as $\dfrac{y^2}{a^2} - \dfrac{x^2}{a^2} = 1$. This is a vertical hyperbola centered at the origin. The foci $F$ and $F'$ are located at $(0, \pm c)$, where $c = \sqrt{a^2 + a^2} = a\sqrt{2}$.

The point $P$ is where the tangent to the hyperbola at $(a, \sqrt{2}a)$ meets the $y$-axis. First, I need to find the equation of the tangent line at this point. Then, I can find the coordinates of $P$. Finally, I need to compute the distances $\overline{PF}$ and $\overline{PF'}$, multiply them, and set the product equal to 8 to solve for $a$.

[/THOUGHT] [PYTHON] from sympy import symbols, Eq, solve, diff, sqrt

a = symbols('a', positive=True) x, y = symbols('x y')

Equation of the hyperbola: y^2/a^2 - x^2/a^2 = 1

hyperbola_eq = Eq(y2/a2 - x2/a2, 1)

Point of tangency: (a, sqrt(2)*a)

point = (a, sqrt(2)*a)

Differentiate implicitly to find the slope of the tangent

dy_dx = diff(hyperbola_eq.lhs, x).subs(hyperbola_eq.rhs, 1) dy_dx = solve(dy_dx, dy_dx[0])[0] # Solve for dy/dx

Slope at the point (a, sqrt(2)*a)

slope = dy_dx.subs({x: point[0], y: point[1]}).simplify()

Equation of the tangent line: y - y1 = m(x - x1)

tangent_line = Eq(y - point[1], slope*(x - point[0]))

Find the y-intercept (P) by setting x = 0

P_y = tangent_line.rhs.subs(x, 0) + point[1] P = (0, P_y)

Foci F and F' are at (0, asqrt(2)) and (0, -asqrt(2))

F = (0, asqrt(2)) F_prime = (0, -asqrt(2))

Distances PF and PF'

PF = sqrt((P[0] - F[0])**2 + (P[1] - F[1])**2) PF_prime = sqrt((P[0] - F_prime[0])**2 + (P[1] - F_prime[1])**2)

Product of distances

product = PF * PF_prime

Set the product equal to 8 and solve for a

equation = Eq(product, 8) solution = solve(equation, a) solution[0].simplify() [/PYTHON]

[THOUGHT] The Python output indicates an error, but the problem can be solved analytically. The hyperbola is $\dfrac{y^2}{a^2} - \dfrac{x^2}{a^2} = 1$, and the tangent at $(a, \sqrt{2}a)$ has slope $1$. The tangent line is $y = x + \sqrt{2}a$, so $P = (0, \sqrt{2}a)$. The foci are at $(0, \pm a\sqrt{2})$. The distances are $\overline{PF} = 2a$ and $\overline{PF'} = 2a$, so their product is $4a^2 = 8$. Thus, $a^2 = 2$, and $p + q = 2 + 1 = 3$. [/THOUGHT]

$\boxed{3}$