Question

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

• A. 18
• B. 22
• C. 26
• D. 34

Hint:

For a hyperbola $e>1$. For an ellipse $0<e<1$. Use the product of roots $e_1 e_2 = 2$ and conditions on the quadratic function $f(x) = x^2 - ax + 2$ like $f(1)$ and the discriminant.

Answer 1

The Correct Option is C

The equation is $x^2 - ax + 2 = 0$. Let its roots be $e_1$ and $e_2$. The product of the roots is $e_1 e_2 = 2$ and the sum is $e_1 + e_2 = a$.

1. Case 1: Both $e_1$ and $e_2$ are eccentricities of hyperbolas. For a hyperbola, eccentricity $e>1$.
Since $e_1>1$ and $e_2>1$, their product $e_1 e_2 = 2>1$ is already satisfied. We also need:
- Real and distinct roots: Discriminant $D>0 \implies a^2 - 8>0 \implies a>2\sqrt{2}$ (since sum $a$ must be positive for $e_1, e_2>1$).
- Both roots greater than 1: For $f(x) = x^2 - ax + 2$, we need $f(1)>0$, the vertex $a/2>1$, and $D>0$.
$f(1) = 1 - a + 2>0 \implies a<3$.
$a/2>1 \implies a>2$.
Combining these, $a \in (2\sqrt{2}, 3)$. Thus, $\alpha = 2\sqrt{2}$ and $\beta = 3$.

2. Case 2: $e_1$ and $e_2$ are eccentricities of an ellipse and a hyperbola.
For an ellipse, $0<e<1$. For a hyperbola, $e>1$.
Let $0<e_1<1$. Since $e_1 e_2 = 2$, it implies $e_2 = 2/e_1$. Since $e_1<1$, $e_2>2$, which satisfies the hyperbola condition ($e_2>1$).
For $f(x)$ to have one root in $(0, 1)$ and one root in $(1, \infty)$, we need $f(1)<0$.
$f(1) = 1 - a + 2<0 \implies a>3$.
Since product is 2 (positive), if one root is positive, both are positive. Roots are real if $a^2 - 8>0$, which is true for $a>3$.
Thus, $a \in (3, \infty)$. This gives $\gamma = 3$.

3. Calculating $\alpha^2 + \beta^2 + \gamma^2$:
$\alpha^2 = (2\sqrt{2})^2 = 8$
$\beta^2 = 3^2 = 9$
$\gamma^2 = 3^2 = 9$
Total sum $= 8 + 9 + 9 = 26$.