The normal to the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{9} = 1$$ at the point $$(8, 3\sqrt{3})$$ on it passes through the point

We need to find the point through which the normal to the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{9} = 1$$ at $$(8, 3\sqrt{3})$$ passes.

Since $$(8, 3\sqrt{3})$$ lies on the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{9} = 1$$, we have $$\frac{64}{a^2} - \frac{27}{9} = 1 \implies \frac{64}{a^2} - 3 = 1 \implies \frac{64}{a^2} = 4 \implies a^2 = 16$$.

Differentiating $$\frac{x^2}{16} - \frac{y^2}{9} = 1$$ implicitly yields $$\frac{2x}{16} - \frac{2y}{9}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{9x}{16y}$$.

At $$(8, 3\sqrt{3})$$ the slope of the tangent is $$\frac{dy}{dx} = \frac{9 \times 8}{16 \times 3\sqrt{3}} = \frac{72}{48\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$$.

Since the slope of the normal is the negative reciprocal, it is $$-\frac{2}{\sqrt{3}}$$, and its equation through $$(8, 3\sqrt{3})$$ is $$y - 3\sqrt{3} = -\frac{2}{\sqrt{3}}(x - 8)$$.

Substituting $$(-1, 9\sqrt{3})$$ into this equation gives $$9\sqrt{3} - 3\sqrt{3} = -\frac{2}{\sqrt{3}}(-1 - 8)$$, or $$6\sqrt{3} = -\frac{2}{\sqrt{3}} \times (-9) = \frac{18}{\sqrt{3}} = 6\sqrt{3}$$, confirming that $$(-1, 9\sqrt{3})$$ lies on the normal.

Therefore, the required point is $$(-1, 9\sqrt{3})$$.