Parametric Coordinates of $$y^2=4ax$$

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