Find Length of Latus Rectum for Locus of Midpoint of Chords

The given equation of the parabola is $y^2 = 4x$. Comparing it with the standard form $y^2 = 4ax$, we have $a = 1$.

The vertex of the parabola is at the origin $O(0,0)$.

Let the parametric coordinates of the points $P$ and $Q$ on the parabola be: $P = (t_1^2, 2t_1)$ $Q = (t_2^2, 2t_2)$

The slope of chord $OP$ ($m_{OP}$) is: $m_{OP} = \frac{2t_1 - 0}{t_1^2 - 0} = \frac{2}{t_1}$

The slope of chord $OQ$ ($m_{OQ}$) is: $m_{OQ} = \frac{2t_2 - 0}{t_2^2 - 0} = \frac{2}{t_2}$

Since $OP$ and $OQ$ are perpendicular to each other ($OP \perp OQ$), the product of their slopes must be $-1$: $m_{OP} \cdot m_{OQ} = -1$ $\left(\frac{2}{t_1}\right) \left(\frac{2}{t_2}\right) = -1 \implies t_1 t_2 = -4$

Let $M(h, k)$ be the mid-point of the line segment $PQ$. By the midpoint formula, we have: $h = \frac{t_1^2 + t_2^2}{2}$ $k = \frac{2t_1 + 2t_2}{2} = t_1 + t_2$

We can express $t_1^2 + t_2^2$ in terms of $(t_1 + t_2)$ and $t_1 t_2$: $t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1 t_2$

Substitute $k = t_1 + t_2$ and $t_1 t_2 = -4$: $t_1^2 + t_2^2 = k^2 - 2(-4) = k^2 + 8$

Now, substitute $t_1^2 + t_2^2 = k^2 + 8$ into the expression for $h$: $h = \frac{k^2 + 8}{2}$ $2h = k^2 + 8$ $k^2 = 2(h - 4)$

Replacing $(h, k)$ with $(x, y)$ gives the locus of the midpoint $M$: $y^2 = 2(x - 4)$

This locus represents a conic $C$, which is a parabola shifted along the x-axis. Comparing $y^2 = 2(x - 4)$ with the standard form $Y^2 = 4AX$, we get: $\text{Length of Latus Rectum} = 4A = 2$

Hence, the length of its latus rectum is 2, which corresponds to Option B.