Problem Analysis #10

To find the minimum possible $x$-coordinate of $R$, we analyze the geometry of the normal lines to the parabola $y = x^2$.

Let $P = (a, a^2)$ with $a > 0$. The slope of the tangent line at $P$ is $2a$, so the slope of the normal line at $P$ is $-\frac{1}{2a}$.

The equation of the normal line at $P$ is:

$$y - a^2 = -\frac{1}{2a}(x - a)$$

To find the intersection of this line with the parabola $y = x^2$, we substitute $y = x^2$:

$$x^2 - a^2 = -\frac{1}{2a}(x - a)$$ $$(x - a)(x + a) = -\frac{1}{2a}(x - a)$$

Since $Q \neq P$, we have $x \neq a$. Dividing both sides by $x - a$ gives the $x$-coordinate of $Q$, which we denote as $b$:

$$b = -a - \frac{1}{2a}$$

Similarly, the normal line at $Q(b, b^2)$ intersects the parabola at $R(c, c^2)$, where the $x$-coordinate $c$ of $R$ is given by:

$$c = -b - \frac{1}{2b}$$

Substituting $b = -\left(a + \frac{1}{2a}\right)$ into the expression for $c$:

$$c = a + \frac{1}{2a} + \frac{1}{2\left(a + \frac{1}{2a}\right)}$$

Let $u = a + \frac{1}{2a}$. Since $a > 0$, by the AM-GM inequality, we have:

$$u \ge 2\sqrt{a \cdot \frac{1}{2a}} = \sqrt{2}$$

We can express $c$ as a function of $u$:

$$c(u) = u + \frac{1}{2u} \quad \text{for } u \ge \sqrt{2}$$

To find the minimum of $c(u)$ on the interval $[\sqrt{2}, \infty)$, we take the derivative with respect to $u$:

$$c'(u) = 1 - \frac{1}{2u^2}$$

For $u \ge \sqrt{2}$, we have $u^2 \ge 2$, which means:

$$c'(u) \ge 1 - \frac{1}{4} = \frac{3}{4} > 0$$

Since the derivative is strictly positive for all $u \ge \sqrt{2}$, the function $c(u)$ is strictly increasing. Thus, the minimum value of $c(u)$ occurs at the boundary $u = \sqrt{2}$:

$$M = c(\sqrt{2}) = \sqrt{2} + \frac{1}{2\sqrt{2}} = \frac{5}{2\sqrt{2}}$$

We square $M$ to find $M^2$:

$$M^2 = \left(\frac{5}{2\sqrt{2}}\right)^2 = \frac{25}{8}$$

This fraction is in lowest terms, so $p = 25$ and $q = 8$.

Finally, we find $p + q$:

$$p + q = 25 + 8 = 33$$

$\boxed{33}$