Let $$P(10, 2\sqrt{15})$$ be a point on the hyperbola $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$, whose foci are S and S'. if the length of its latus rectum is 8, then the square of the area of $$\Delta PSS'$$ is equal to:

$$\frac{2b^{2}}{a}=8 \;\Longrightarrow\; \frac{b^{2}}{a}=4 \;\Longrightarrow\; b^{2}=4a \;$$ $$-(1)$$

The point $$P(10,\,2\sqrt{15})$$ lies on the hyperbola, so it satisfies the equation:
$$\frac{10^{2}}{a^{2}}-\frac{(2\sqrt{15})^{2}}{b^{2}} = 1$$
$$\Longrightarrow \frac{100}{a^{2}}-\frac{60}{b^{2}} = 1$$ $$-(2)$$

From $$(1)$$: $$a = \frac{b^{2}}{4} \;\Longrightarrow\; a^{2}=\frac{b^{4}}{16}$$.

Substituting $$a^{2}$$ from above and $$b^{2}$$ itself into $$(2)$$:
$$\frac{100}{\,b^{4}/16\,} - \frac{60}{b^{2}} = 1$$
$$\Longrightarrow \frac{1600}{b^{4}} - \frac{60}{b^{2}} - 1 = 0$$.

Put $$t = \frac{1}{b^{2}}$$. The quadratic becomes
$$1600t^{2}-60t-1=0$$.

Solving, the discriminant is $$\Delta = (-60)^{2}-4\cdot1600(-1)=10000$$, hence
$$t = \frac{60\pm100}{3200}\;.$$

Only the positive root is acceptable: $$t = \frac{160}{3200}=\frac{1}{20}$$, so
$$b^{2} = 20$$.

From $$(1)$$, $$a = \frac{b^{2}}{4} = \frac{20}{4}=5$$.

Hence $$c^{2}=a^{2}+b^{2}=25+20=45 \;\Longrightarrow\; c=\sqrt{45}=3\sqrt{5}$$.

The base $$SS'$$ of $$\triangle PSS'$$ lies on the x-axis, so its length is $$2c$$.
The perpendicular height from $$P$$ to this base equals the y-coordinate of $$P$$, namely $$2\sqrt{15}$$.

 the area of the triangle is
$$\text{Area}= \frac12 \times (2c)\times(2\sqrt{15}) = 2c\sqrt{15}$$.
$$\bigl(\text{Area}\bigr)^{2}= \bigl(2c\sqrt{15}\bigr)^{2}=4c^{2}\times15 = 60c^{2}$$.

Using $$c^{2}=45$$ we get
$$60c^{2}=60\times45=2700$$.