Problem Analysis #45

[THOUGHT] Given the hyperbola equation (\dfrac{x^2}{a^2} - \dfrac{y^2}{a^2} = -1), this rewrites as (\dfrac{y^2}{a^2} - \dfrac{x^2}{a^2} = 1), which is a standard form for a hyperbola centered at the origin with a vertical transverse axis.

First, let's find the coordinates of the foci (F) and (F'). For a hyperbola of the form (\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1), the distance from the center to the foci is given by (c), where (c = \sqrt{a^2 + b^2}). Here, (a^2 = a^2) and (b^2 = a^2), so: [ c = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}. ] Thus, the foci are at ((0, a\sqrt{2})) and ((0, -a\sqrt{2})).

Next, we find the equation of the tangent to the hyperbola at the point ((a, \sqrt{2}a)). The equation of the tangent to the hyperbola (\dfrac{y^2}{a^2} - \dfrac{x^2}{a^2} = 1) at the point ((x_1, y_1)) is given by: [ \dfrac{y_1 y}{a^2} - \dfrac{x_1 x}{a^2} = 1. ] Substituting ((x_1, y_1) = (a, \sqrt{2}a)), we get: [ \dfrac{\sqrt{2}a \cdot y}{a^2} - \dfrac{a \cdot x}{a^2} = 1, ] which simplifies to: [ \dfrac{\sqrt{2} y}{a} - \dfrac{x}{a} = 1. ] Multiplying through by (a), we obtain: [ \sqrt{2} y - x = a. ]

To find the y-intercept (P) of this tangent line, set (x = 0): [ \sqrt{2} y = a \implies y = \dfrac{a}{\sqrt{2}} = \dfrac{a\sqrt{2}}{2}. ] Thus, (P) is the point ((0, \dfrac{a\sqrt{2}}{2})).

Now, we calculate the distances (\overline{PF}) and (\overline{PF'}): [ \overline{PF} = \left| \dfrac{a\sqrt{2}}{2} - a\sqrt{2} \right| = \left| \dfrac{a\sqrt{2} - 2a\sqrt{2}}{2} \right| = \left| \dfrac{-a\sqrt{2}}{2} \right| = \dfrac{a\sqrt{2}}{2}, ] [ \overline{PF'} = \left| \dfrac{a\sqrt{2}}{2} - (-a\sqrt{2}) \right| = \left| \dfrac{a\sqrt{2}}{2} + a\sqrt{2} \right| = \left| \dfrac{a\sqrt{2} + 2a\sqrt{2}}{2} \right| = \left| \dfrac{3a\sqrt{2}}{2} \right| = \dfrac{3a\sqrt{2}}{2}. ]

According to the problem, (\overline{PF} \times \overline{PF'} = 8): [ \left( \dfrac{a\sqrt{2}}{2} \right) \times \left( \dfrac{3a\sqrt{2}}{2} \right) = 8. ] Simplifying the left-hand side: [ \dfrac{a\sqrt{2} \cdot 3a\sqrt{2}}{4} = \dfrac{3a^2 \cdot 2}{4} = \dfrac{6a^2}{4} = \dfrac{3a^2}{2}. ] Setting this equal to 8: [ \dfrac{3a^2}{2} = 8. ] Solving for (a^2): [ 3a^2 = 16 \implies a^2 = \dfrac{16}{3}. ]

Given (a^2 = \dfrac{16}{3}), we need (p + q) where (a^2 = \dfrac{p}{q}) and (\gcd(p, q) = 1). Here, (p = 16) and (q = 3), so: [ p + q = 16 + 3 = 19. ]

Thus, the final answer is: [ \boxed{19} ] [/THOUGHT]