Let e_1 and e_2 be two distinct roots of x^2-ax+2=0.

Question

Let $e_1$ and $e_2$ be two distinct roots of $x^2-ax+2=0$. Let $\{a\in\mathbb R: e_1,e_2 	ext{ are both eccentricities of hyperbolas}\}=(\alpha,\beta)$, and $\{a\in\mathbb R: e_1,e_2 	ext{ are the eccentricities of an ellipse and a hyperbola respectively}\}=(\gamma,\infty)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to:

Accepted Answer

By Vieta: $e_1+e_2=a,\ e_1e_2=2$; for distinct real roots, $a^2-8>0\Rightarrow |a|>2\sqrt2$. Both roots $>1$ (hyperbola-hyperbola): need $f(1)=3-a>0$, axis of symmetry $\dfrac a2>1$, and discriminant $>0$. This gives $a<3$, $a>2$, $a>2\sqrt2$, combining to $a\in(2\sqrt2,3)$. So $\alpha=2\sqrt2,\ \beta=3$. One root in $(0,1)$, other $>1$ (ellipse-hyperbola): since $e_1e_2=2>0$ both roots are positive, so we need $f(1)<0$: $3-a<0\Rightarrow a>3$, together with $a^2-8>0$ (automatic). So $\gamma=3$, set $=(3,\infty)$. $$\alpha^2+\beta^2+\gamma^2 = 8+9+9=26.$$

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