The shortest distance between the curves $$y^2 = 8x$$ and $$x^2 + y^2 + 12y + 35 = 0$$ is:

Let the given parabola be:

$$y^2 = 8x$$

Comparing with the standard form $$y^2 = 4ax$$, we get $$4a = 8 \implies a = 2$$. Any parametric point on this parabola can be represented as $$P(2t^2, 4t)$$.

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Let the given circle be:

$$x^2 + y^2 + 12y + 35 = 0$$

By completing the square, we can rewrite the equation of the circle as:

$$x^2 + (y + 6)^2 - 36 + 35 = 0 \implies x^2 + (y + 6)^2 = 1$$

Thus, the center of the circle is $$C(0, -6)$$ and its radius is $$R = 1$$.

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The shortest distance between a parabola and a circle occurs along the common normal line, which must pass through the center of the circle $$C(0, -6)$$.

The equation of the normal to the parabola $$y^2 = 4ax$$ at the parametric point $$t$$ is:

$$y = -tx + 2at + at^3$$

Substituting $$a = 2$$ into the normal equation:

$$y = -tx + 4t + 2t^3$$

Since this normal passes through the center of the circle $$(0, -6)$$, substituting these coordinates gives:

$$-6 = -t(0) + 4t + 2t^3$$

$$2t^3 + 4t + 6 = 0 \implies t^3 + 2t + 3 = 0$$

Testing for integer roots, we find that $$t = -1$$ satisfies the equation because $$(-1)^3 + 2(-1) + 3 = 0$$. Since the derivative of the polynomial expression $$3t^2 + 2$$ is always positive, $$t = -1$$ is the only real root.

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Substituting $$t = -1$$ back into the parametric coordinates of point $$P$$:

$$P = (2(-1)^2, 4(-1)) = (2, -4)$$

Now, we calculate the distance between the point $$P(2, -4)$$ on the parabola and the center of the circle $$C(0, -6)$$:

$$PC = \sqrt{(2 - 0)^2 + (-4 - (-6))^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$$

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The shortest distance between the two curves is obtained by subtracting the radius of the circle from this distance:

$$\text{Shortest Distance} = PC - R = 2\sqrt{2} - 1$$

Therefore, the shortest distance between the curves is equal to $$2\sqrt{2} - 1$$.