Let e_1 and e_2 be two distinct roots of the | vedclass

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(C) For a hyperbola, eccentricity $e > 1$. For an ellipse, $0 < e < 1$. The roots of $x^2 - ax + 2 = 0$ are $e_1, e_2$.
Sum $e_1 + e_2 = a$

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(C) For a hyperbola, eccentricity $e > 1$. For an ellipse, $0 < e < 1$. The roots of $x^2 - ax + 2 = 0$ are $e_1, e_2$. Sum $e_1 + e_2 = a$ and product $e_1 e_2 = 2$. For $S_1$, both $e_1, e_2 > 1$. Since $e_1 e_2 = 2$, if $e_1 > 1$, then $e_2 = 2/e_1 < 2$. Also, $D = a^2 - 8 > 0 \implies a > 2\sqrt{2} \approx 2.828$. For $e_1, e_2 > 1$, let $f(x) = x^2 - ax + 2$. We need $f(1) > 0$ and vertex $a/2 > 1$. $f(1) = 1 - a + 2 = 3 - a > 0 \implies a < 3$. Thus, $S_1 = (2\sqrt{2}, 3)$, so $\alpha = 2\sqrt{2}$ and $\beta = 3$. For $S_2$, one root $e_1 < 1$ and other $e_2 > 1$. This implies $f(1) < 0 \implies 3 - a < 0 \implies a > 3$. Thus, $\gamma = 3$. We need $\alpha^2 + \beta^2 + \gamma^2 = (2\sqrt{2})^2 + 3^2 + 3^2 = 8 + 9 + 9 = 26$.

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