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]