The equation of the parabola is given as $$y^2 = 4x$$. Comparing this with the standard equation $$y^2 = 4ax$$, we find that $$a = 1$$.
The parametric coordinates of any arbitrary point on this parabola can be written in the form $$(at^2, 2at) = (t^2, 2t)$$.
Let us define the coordinates for points $$P$$ and $$Q$$ on the parabola using parameters $$t_1$$ and $$t_2$$:
- $$P = (t_1^2, 2t_1)$$
- $$Q = (t_2^2, 2t_2)$$
- The vertex is the origin, $$O = (0, 0)$$.
We are given that the chords $$OP$$ and $$OQ$$ are at right angles ($$OP \perp OQ$$). This means the product of their slopes must equal $$-1$$:
$$\text{Slope of } OP \, (m_1) = \frac{2t_1 - 0}{t_1^2 - 0} = \frac{2}{t_1}$$ $$\text{Slope of } OQ \, (m_2) = \frac{2t_2 - 0}{t_2^2 - 0} = \frac{2}{t_2}$$
Using the perpendicular condition ($$m_1 \cdot m_2 = -1$$):
$$\left(\frac{2}{t_1}\right) \cdot \left(\frac{2}{t_2}\right) = -1$$ $$\frac{4}{t_1t_2} = -1 \implies t_1t_2 = -4$$
The equation of a straight line passing through two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is:
$$y - y_1 = \left(\frac{y_2 - y_1}{x_2 - x_1}\right)(x - x_1)$$
Substituting the coordinates of points $$P$$ and $$Q$$:
$$y - 2t_1 = \left(\frac{2t_2 - 2t_1}{t_2^2 - t_1^2}\right)(x - t_1^2)$$
We can simplify the slope term by factoring the denominator:
$$\frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_1 + t_2}$$
Now, substitute this simplified slope back into the line equation:
$$y - 2t_1 = \frac{2}{t_1 + t_2}(x - t_1^2)$$
Multiply both sides by $(t_1 + t_2)$ to clear the fraction:
$$(t_1 + t_2)y - 2t_1(t_1 + t_2) = 2x - 2t_1^2$$ $$(t_1 + t_2)y - 2t_1^2 - 2t_1t_2 = 2x - 2t_1^2$$
Cancel out the common $$-2t_1^2$$ term from both sides:
$$(t_1 + t_2)y - 2t_1t_2 = 2x$$
Now, substitute our relation $$t_1t_2 = -4$$ from Step 2 into the chord equation:
$$(t_1 + t_2)y - 2(-4) = 2x$$ $$(t_1 + t_2)y + 8 = 2x$$
The axis of the parabola $$y^2 = 4x$$ is the x-axis, where the equation of the line is $$y = 0$$.
To find where the chord $$PQ$$ intersects this axis, substitute $$y = 0$$ into our equation:
$$(t_1 + t_2)(0) + 8 = 2x$$ $$8 = 2x \implies x = 4$$
Since the value of $$x$$ is constant ($$4$$) and entirely independent of the variables $$t_1$$ and $$t_2$$, the chord $$PQ$$ will always intersect the axis of the parabola at the fixed coordinate point $$(4, 0)$$.
This matches Option B
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