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

We are dealing with a quadratic equation $x^2 - ax + 2 = 0$ whose roots $e_1, e_2$ represent conic section eccentricities. We know that eccentricity $e$ for an ellipse is in $(0, 1)$ and for a hyperbola is in $(1, \infty)$.

First, analyze the condition for both roots to be greater than 1 (hyperbolas).
For roots $e_1, e_2>1$:
i) Roots must be real: $D = a^2 - 8>0 \implies a>2\sqrt{2}$ (as $a=e_1+e_2>0$).
ii) $(e_1-1)(e_2-1)>0 \implies e_1 e_2 - (e_1+e_2) + 1>0 \implies 2 - a + 1>0 \implies a<3$.
Combining these, $a \in (2\sqrt{2}, 3)$, so $\alpha = 2\sqrt{2}, \beta = 3$.

Second, analyze the condition for one root in $(0, 1)$ and the other in $(1, \infty)$.
Since the product $e_1 e_2 = 2$, if $e_1 \in (0, 1)$, then $e_2 = 2/e_1$ must be greater than 2. Thus the condition of one ellipse and one hyperbola is satisfied if one root is between 0 and 1.
For a quadratic $f(x) = x^2 - ax + 2$, the roots lie on either side of $x=1$ if $f(1)<0$.
$1^2 - a(1) + 2<0 \implies 3 - a<0 \implies a>3$.
The range is $(3, \infty)$, so $\gamma = 3$.

Finally, evaluate the required sum:
$\alpha^2 + \beta^2 + \gamma^2 = (2\sqrt{2})^2 + 3^2 + 3^2 = 8 + 9 + 9 = 26$.