Let a tangent to the curve $$y^2 = 24x$$ meet the curve $$xy = 2$$ at the points $$A$$ and $$B$$. Then the mid-points of such line segments $$AB$$ lie on a parabola with the

The parabola $$y^2 = 24x$$ has $$4a = 24$$, so $$a = 6$$. A parametric point is $$(6t^2, 12t)$$ and the tangent at this point is $$ty = x + 6t^2$$.

This tangent meets the rectangular hyperbola $$xy = 2$$. Substituting $$x = ty - 6t^2$$:

$$(ty - 6t^2)y = 2 \Rightarrow ty^2 - 6t^2 y - 2 = 0$$

Let the y-coordinates of points A and B be $$y_1, y_2$$. By Vieta's formulas:

$$y_1 + y_2 = 6t, \quad y_1 y_2 = \frac{-2}{t}$$

The midpoint coordinates are:

$$k = \frac{y_1 + y_2}{2} = 3t \Rightarrow t = \frac{k}{3}$$

$$h = \frac{x_1 + x_2}{2} = \frac{1}{2}\left(\frac{2}{y_1} + \frac{2}{y_2}\right) = \frac{y_1 + y_2}{y_1 y_2} = \frac{6t}{-2/t} = -3t^2$$

Substituting $$t = k/3$$: $$h = -3 \cdot \frac{k^2}{9} = -\frac{k^2}{3}$$

The locus is $$y^2 = -3x$$, a parabola of the form $$y^2 = -4bx$$ with $$4b = 3$$, so $$b = \frac{3}{4}$$.

The directrix of $$y^2 = -3x$$ is $$x = \frac{3}{4}$$, i.e., $$4x = 3$$.

The length of the latus rectum is $$4b = 3$$.

The answer is Option A: directrix $$4x = 3$$.