Let the tangent to the curve $$x^2 + 2x - 4y + 9 = 0$$ at the point P(1, 3) on it meet the y-axis at A. Let the line passing through P and parallel to the line $$x - 3y = 6$$ meet the parabola $$y^2 = 4x$$ at B. If B lies on the line $$2x - 3y = 8$$, then $$AB^2$$ is equal to ______.

The curve is $$x^2 + 2x - 4y + 9 = 0$$, i.e., $$y = \frac{x^2+2x+9}{4}$$.

First, we find tangent at P(1,3).

$$y' = \frac{2x+2}{4} = \frac{x+1}{2}$$. At $$x=1$$: $$y' = 1$$.

Tangent: $$y - 3 = 1(x-1) \implies y = x + 2$$.

Meets y-axis at A(0, 2).

Next, we find B on parabola $$y^2 = 4x$$.

Line through P(1,3) parallel to $$x-3y=6$$ has slope $$\frac{1}{3}$$:

$$y - 3 = \frac{1}{3}(x-1) \implies y = \frac{x+8}{3}$$

Substituting into $$y^2 = 4x$$:

$$\frac{(x+8)^2}{9} = 4x \implies x^2+16x+64 = 36x \implies x^2-20x+64 = 0$$

$$x = \frac{20 \pm 12}{2} = 16 \text{ or } 4$$

B must lie on $$2x-3y=8$$:

At $$x=16$$: $$y = 8$$, check: $$2(16)-3(8) = 8$$ ✓. So B = (16, 8).

From this, we compute AB².

$$AB^2 = (16-0)^2 + (8-2)^2 = 256 + 36 = 292$$

The answer is 292.