Let O be the vertex of the parabola $$y^2 = 4x$$ and its chords OP and OQ are perpendicular to each other. If the locus of the mid-point of the line segment PQ is a conic C, then the length of its latus rectum is :

To solve this, we will use parametric coordinates for the points on the parabola $$y^2 = 4x$$.

1. Identify Points P and Q

For the parabola $$y^2 = 4ax$$ (where $$a=1$$), any point can be represented as $$(at^2, 2at) = (t^2, 2t)$$.

• Let $$P = (t_1^2, 2t_1)$$
• Let $$Q = (t_2^2, 2t_2)$$
• The vertex is $$O = (0,0)$$.
• We know $$(t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2$$
• Substitute our values: $$k^2 = 2h + 2(-4)$$
• $$k^2 = 2h - 8 \implies k^2 = 2(h - 4)$$
• Comparing $$y^2 = 2(x - 4)$$ to $$Y^2 = 4AX$$:
• $$4A = 2$$

2. Use the Perpendicular Condition

The chords $$OP$$ and $$PQ$$ are perpendicular, so the product of their slopes is $$-1$$:

$$m_{OP} \cdot m_{OQ} = -1$$

$$\left( \frac{2t_1}{t_1^2} \right) \cdot \left( \frac{2t_2}{t_2^2} \right) = -1$$

$$\frac{4}{t_1 t_2} = -1 \implies \mathbf{t_1 t_2 = -4}$$

3. Find the Locus of the Mid-point

Let the mid-point of $$PQ$$ be $$(h, k)$$.

$$h = \frac{t_1^2 + t_2^2}{2} \quad \text{and} \quad k = \frac{2t_1 + 2t_2}{2} = t_1 + t_2$$

We need to eliminate $$t_1$$ and $$t_2$$ to find the relationship between $$h$$ and $$k$$:

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

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

4. Determine the Latus Rectum

The equation $$y^2 = 2(x - 4)$$ is a parabola in the standard form $$Y^2 = 4AX$$, where $$4A$$ is the length of the latus rectum.

The length of the latus rectum is 2.

Correct Option: B