Find Value of p Squared Plus q Squared for Parabola Minimum Vertex Distance

To find the value of $p^2 + q^2$, we follow these steps:

1. Utilize the given point on the parabola: The equation of the parabola is: $y = x^2 + px + q$ Since the parabola passes through the point $(1, -1)$, we substitute $x = 1$ and $y = -1$: $-1 = (1)^2 + p(1) + q \implies 1 + p + q = -1 \implies q = -p - 2$

2. Find the vertex of the parabola: The $x$-coordinate of the vertex of a parabola $y = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. Here $a = 1$ and $b = p$, so: $x_v = -\frac{p}{2}$

   Substituting $x_v$ back into the parabola equation gives the $y$-coordinate of the vertex ($y_v$): $y_v = \left(-\frac{p}{2}\right)^2 + p\left(-\frac{p}{2}\right) + q = \frac{p^2}{4} - \frac{p^2}{2} + q = q - \frac{p^2}{4}$

3. Express the distance from the vertex to the $x$-axis: Substitute $q = -p - 2$ into $y_v$: $y_v = -p - 2 - \frac{p^2}{4} = -\frac{1}{4}(p^2 + 4p + 8) = -\frac{1}{4}\left[(p+2)^2 + 4\right]$

   Since $(p+2)^2 \ge 0$, the quantity inside the brackets is always positive ($\ge 4$), which means $y_v \le -1$ for all real values of $p$.

   The distance $d$ between the vertex and the $x$-axis is given by $|y_v|$: $d = |y_v| = \frac{1}{4}(p+2)^2 + 1$

4. Minimize the distance $d$: The distance $d$ is minimized when $(p + 2)^2 = 0$, which occurs when: $p = -2$

   Using $q = -p - 2$, we find $q$: $q = -(-2) - 2 = 0$

5. Calculate $p^2 + q^2$: $p^2 + q^2 = (-2)^2 + 0^2 = 4$

Thus, the correct option is B.