Let the parabola $$y = x^2 + px + q$$ passing through the point $$(1, -1)$$ be such that the distance between its vertex and the x-axis is minimum. Then the value of $$p^2 + q^2$$ is :

The general equation of the parabola is $$y = x^2 + px + q$$.

Because the parabola passes through $$(1,-1)$$, substitute $$x = 1,\, y = -1$$:

$$-1 = 1 + p + q \;\;\Longrightarrow\;\; q = -2 - p \;-(1)$$

The vertex of $$y = x^2 + px + q$$ lies at

$$x_v = -\frac{p}{2}, \qquad y_v = -\frac{p^2}{4} + q \;-(2)$$

Using $$(1)$$ in $$(2)$$, express the y-coordinate of the vertex only in terms of $$p$$:

$$y_v = -\frac{p^2}{4} + (-2 - p) = -\frac{p^2}{4} - p - 2 = -\frac{p^2 + 4p + 8}{4}$$

Distance of the vertex from the x-axis is the absolute value of $$y_v$$:

$$D(p) = \left| -\frac{p^2 + 4p + 8}{4} \right| = \frac{p^2 + 4p + 8}{4}$$

Minimising $$D(p)$$ is equivalent to minimising the quadratic numerator

$$f(p) = p^2 + 4p + 8$$

The minimum of a quadratic $$ap^2 + bp + c$$ occurs at $$p = -\frac{b}{2a}$$. Here, $$a = 1,\; b = 4$$, so

$$p_{\text{min}} = -\frac{4}{2} = -2$$

With $$p = -2$$, calculate $$q$$ from $$(1)$$:

$$q = -2 - (-2) = 0$$

Finally, evaluate $$p^2 + q^2$$:

$$p^2 + q^2 = (-2)^2 + 0^2 = 4$$

Therefore, the required value is $$4$$.

Option B which is: $$4$$