Let $e_{1}$ and $e_{2}$ be two distinct roots of the equation $x^{2}-ax+2=0$. Let the sets $\{a \in \mathbb{R}: e_{1}$ and $e_{2}$ are the eccentricities of hyperbolas $\}=(\alpha, \beta)$, and $\{a \in \mathbb{R}: e_{1}$ and $e_{2}$ are the eccentricities of an ellipse and a hyperbola, respectively $\}=(\gamma, \infty)$. The $\alpha^{2}+\beta^{2}+\gamma^{2}$ is equal to:

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Quadratic Equations and Inequalities

Let

e_{1}

and

e_{2}

be two distinct roots of the equation

x^{2} - a x + 2 = 0

. Let the sets

{a \in R : e_{1}

and

e_{2}

are the eccentricities of hyperbolas

} = (α, β)

, and

{a \in R : e_{1}

and

e_{2}

are the eccentricities of an ellipse and a hyperbola, respectively

} = (γ, \infty)

.
The

α^{2} + β^{2} + γ^{2}

is equal to:

Answer: (C)

The roots satisfy

e_{1} + e_{2} = a

and

e_{1} e_{2} = 2

, so

e_{2} = \frac{2}{e _{1}}

. For both to be eccentricities of hyperbolas,

e_{1} > 1

and

\frac{2}{e _{1}} > 1

, i.e.

e_{1} \in (1, 2)

. On this interval

a = e_{1} + \frac{2}{e _{1}}

has least value

22

e_{1} = 2

and approaches

3

at both ends, so

a \in (22, 3)

, giving

α = 22

and

β = 3

. For an ellipse and a hyperbola,

0 < e_{1} < 1

with

e_{2} = \frac{2}{e _{1}} > 1

automatically; as

e_{1} \to 1^{-}

a \to 3

and as

e_{1} \to 0^{+}

a \to \infty

, so

a \in (3, \infty)

and

γ = 3

. Therefore

α^{2} + β^{2} + γ^{2} = 8 + 9 + 9 = 26

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option