Find Sum of Squares of Boundaries for Eccentricity Sets

Given the quadratic equation: $x^2 - ax + 2 = 0$

Let $e_1$ and $e_2$ be its roots (with $e_1 < e_2$). By Vieta's formulas, we have: $e_1 + e_2 = a \quad \text{and} \quad e_1 e_2 = 2$

Since $e_1$ and $e_2$ represent eccentricities, they must be real and positive. The discriminant $D$ of the quadratic equation must be positive for distinct real roots: $D = a^2 - 4(1)(2) = a^2 - 8 > 0 \implies a > 2\sqrt{2} \quad (\text{since } e_1 + e_2 = a > 0)$

The function relating the smaller root $e_1$ to $a$ is: $a = e_1 + e_2 = e_1 + \frac{2}{e_1}$

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Case 1: Both $e_1$ and $e_2$ are eccentricities of hyperbolas

For a hyperbola, the eccentricity $e > 1$. Thus, we require $1 < e_1 < e_2$.

Since $e_1 e_2 = 2$ and $e_1 < e_2$, we must have: $1 < e_1 < \sqrt{2} < e_2 < 2$

Consider the continuous function $f(x) = x + \frac{2}{x}$ for $x \in (1, \sqrt{2})$:

• As $x \to 1^+$, $f(x) \to 1 + 2 = 3$.
• As $x \to \sqrt{2}^-$, $f(x) \to \sqrt{2} + \sqrt{2} = 2\sqrt{2}$.

Since $f(x)$ is strictly decreasing on $(1, \sqrt{2})$, the range of $a$ is: $a \in (2\sqrt{2}, 3)$

Comparing this with the given set $(\alpha, \beta)$, we get: $\alpha = 2\sqrt{2} \quad \text{and} \quad \beta = 3$

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Case 2: $e_1$ and $e_2$ are eccentricities of an ellipse and a hyperbola, respectively

For an ellipse, $0 < e_1 < 1$, and for a hyperbola, $e_2 > 1$.

Since $e_1 e_2 = 2$ and $0 < e_1 < 1$, it follows that $e_2 = \frac{2}{e_1} > 2$.

Again, using $a = e_1 + \frac{2}{e_1}$ for $e_1 \in (0, 1)$:

• As $e_1 \to 1^-$, $a \to 1 + 2 = 3$.
• As $e_1 \to 0^+$, $a \to \infty$.

Thus, the range of $a$ is: $a \in (3, \infty)$

Comparing this with the given set $(\gamma, \infty)$, we get: $\gamma = 3$

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Calculating $\alpha^2 + \beta^2 + \gamma^2$:

$\alpha^2 = (2\sqrt{2})^2 = 8$ $\beta^2 = 3^2 = 9$ $\gamma^2 = 3^2 = 9$

Therefore: $\alpha^2 + \beta^2 + \gamma^2 = 8 + 9 + 9 = 26$