A particle is moving in the $$xy$$-plane along a curve $$C$$ passing through the point $$(3, 3)$$. The tangent to the curve $$C$$ at the point $$P$$ meets the $$x$$-axis at $$Q$$. If the $$y$$-axis bisects the segment $$PQ$$, then $$C$$ is a parabola with

Let $$P(x, y)$$ be a point on the curve $$C$$. The tangent at $$P$$ has slope $$\frac{dy}{dx}$$.

The equation of the tangent at $$P$$ is $$Y - y = \frac{dy}{dx}(X - x)$$. It meets the x-axis where $$Y = 0$$, so $$0 - y = \frac{dy}{dx}(X - x) \implies X = x - \frac{y}{dy/dx}$$, and thus $$Q = \left(x - \frac{y}{y'}, 0\right)$$.

The midpoint of $$PQ$$ is $$M = \left(\frac{x + x - \frac{y}{y'}}{2}, \frac{y}{2}\right)$$. Since the y-axis bisects $$PQ$$, this midpoint lies on the y-axis, giving $$\frac{2x - \frac{y}{y'}}{2} = 0 \implies 2x = \frac{y}{y'} \implies y' = \frac{y}{2x}$$.

Solving the differential equation $$\frac{dy}{dx} = \frac{y}{2x}$$ gives $$\frac{dy}{y} = \frac{dx}{2x}$$, $$\ln|y| = \frac{1}{2}\ln|x| + C$$, and hence $$y^2 = kx$$.

Since the curve passes through $$(3, 3)$$, we have $$9 = 3k \implies k = 3$$. Therefore, the curve is $$y^2 = 3x$$, a parabola of the form $$y^2 = 4ax$$ with $$4a = 3$$, so the length of the latus rectum is $$4a = 3$$.

Therefore, the answer is Option A: length of latus rectum 3.