Co-Normal Parabolic Centroid

Comparing the equation $y^2 = 8x$ with $y^2 = 4ax$ , we get $a=2$

Let the points be $A(2t_1^2, 4t_1), B(2t_2^2, 4t_2), C(2t_3^2, 4t_3)$ . Let the points $A \text{ and } B$ lie above x-axis. As the points are co-normal, so $t_1 + t_2 + t_3 = 0$

The normals at the points are concurrent at $M(h, k) \equiv M(2a + a(t_1^2 + t_2^2 + t_1t_2), -at_1t_2(t_1 + t_2))$ . The centroid of the triangle is $\displaystyle G\bigg(\frac{a}{3}(t_1^2 + t_2^2 + t_3^2), 0\bigg)$ .

Comparing, with centroid, we get

$\frac{a}{3}(t_1^2 + t_2^2 + t_3^2) = 4$

$\frac23(t_1^2 + t_2^2 + (-t_1 -t_2)^2) = 4...... (\text{As } t_1 + t_2 + t_3 = 0)$

$\boxed{t_1^2 + t_2^2 + t_1t_2 = 3}$

---

So,

$h = 2a + a(t_1^2 + t_2^2 + t_1t_2)$

$\boxed{h = 10}$

Now, applying AM-GM inequality , we get

$\frac{t_1^2 + t_2^2 + t_1t_2}{3} \geq \sqrt[3]{t_1^3t_2^3}$

$1 \geq t_1t_2$

$\boxed{t_1t_2 \leq 1} \ldots (1)$

---

Now,

$t_1^2 + t_2^2 + t_1t_2 = 3$

$(t_1 + t_2)^2 = 3 + t_1t_2 \leq 3 + 1 \ldots (\text{By } (1))$

$(t_1 + t_2)^2 \leq 4$

$\boxed{t_1 + t_2 \leq 2} \ldots (2)$

By $(1)$ and $(2)$

$t_1t_2(t_1 + t_2) \leq 2$ $k = -at_1t_2(t_1 + t_2)$ $\boxed{k \gt -4}$

---

Therefore, $h = a = 10 \text{ and } k \gt -4 = b$

So, $\color{#3D99F6}{\boxed{a^2 + b^2 = 116}}$

The equation of the normal to the parabola at the point $(2t^2,4t)$ is $tx + y = 4t + 2t^3$ . Thus $M(h,k)$ lies on three normals provided that the cubic equation $f(t) \; = \; t^3 + (2 - \tfrac12h)t - \tfrac12k \; = \; 0$ has three real roots. It these three roots are $t_1,t_2,t_3$ , then $t_1 + t_2 + t_3 = 0$ and $t_1t_2 + t_1t_3 + t_2t_3 = 2 - \tfrac12h$ , and hence $t_1^2+t_2^2+t_3^2 = h - 4$ . Thus the centroid of $A,B,C$ is $G$ with coordinates $\left(\tfrac23(t_1^2+t_2^2+t_3^2),\tfrac43(t_1+t_2+t_3)\right) \; = \; \left(\tfrac23(h-4),0\right)$ and hence we deduce that $h=10$ . Thus $f(t) = t^3 - 3t - \tfrac12k$ has turning points at $\pm1$ .

For $f(t)$ to have at least two non-negative roots, we need $f(0) \ge 0 > f(1)$ , which implies that $-4 < k \le 0$ . Thus $a=10$ and $b = -4$ , making the answer $\boxed{116}$ .