Let O be the vertex of the parabola y^2=4x and its chords OP and OQ...

Let the vertex of the parabola $y^2 = 4x$ be $O(0, 0)$.

Let the coordinates of the points $P$ and $Q$ on the parabola be $(t_1^2, 2t_1)$ and $(t_2^2, 2t_2)$ respectively.

The slope of the chord $OP$ is $m_1 = \dfrac{2t_1 - 0}{t_1^2 - 0} = \dfrac{2}{t_1}$.

The slope of the chord $OQ$ is $m_2 = \dfrac{2t_2 - 0}{t_2^2 - 0} = \dfrac{2}{t_2}$.

Since $OP$ and $OQ$ are perpendicular to each other, $m_1 m_2 = -1$.

$\Rightarrow \left(\dfrac{2}{t_1}\right) \left(\dfrac{2}{t_2}\right) = -1$

$\Rightarrow t_1 t_2 = -4$

Let $M(h, k)$ be the mid-point of the line segment $PQ$. Then,

$h = \dfrac{t_1^2 + t_2^2}{2} \Rightarrow t_1^2 + t_2^2 = 2h$

$k = \dfrac{2t_1 + 2t_2}{2} \Rightarrow t_1 + t_2 = k$

Squaring the equation for $k$, we get:

$k^2 = (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1 t_2$

Substituting the values of $t_1^2 + t_2^2$ and $t_1 t_2$, we obtain:

$k^2 = 2h + 2(-4)$

$k^2 = 2h - 8$

$k^2 = 2(h - 4)$

Replacing $h$ with $x$ and $k$ with $y$, the locus of the mid-point is:

$y^2 = 2(x - 4)$

This represents a parabola of the form $Y^2 = 4AX$, where the length of the latus rectum is $4A = 2$.

Answer: $2$