JEE Main 2026 Apr 6 Shift 1, Mathematics Q3: Roots Lie In An Interval

JEE Main2026Apr 6, Shift 1Mathematics

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 ${a \in R : e_{1} and e_{2} are the eccentricities of an ellipse and a hyperbola, respectively} = (γ, \infty)$ .

Then $α^{2} + β^{2} + γ^{2}$ is equal to:

A
$18$
B
$22$
C
$26$
D
$34$

Solution

From Vieta's: $e_{1} + e_{2} = a$ and $e_{1} e_{2} = 2$ , so $e_{2} = \frac{2}{e _{1}}$ and $a = e_{1} + \frac{2}{e _{1}}$ .

Case 1 (both hyperbolas): require $e_{1} > 1$ and $e_{2} > 1$ . From $e_{2} = 2/ e_{1} > 1$ we get $e_{1} < 2$ , so $e_{1} \in (1, 2)$ .

On this interval $a = e_{1} + 2/ e_{1}$ attains its minimum at $e_{1} = 2$ , giving $a_{m i n} = 22$ , and approaches $3$ at either endpoint. Hence $a \in (22, 3)$ , so $α = 22$ , $β = 3$ .

Case 2 (ellipse and hyperbola): $0 < e_{1} < 1$ and $e_{2} > 1$ . Here $a = e_{1} + 2/ e_{1}$ with $e_{1} \in (0, 1)$ : as $e_{1} \to 1^{-}$ , $a \to 3$ , and as $e_{1} \to 0^{+}$ , $a \to \infty$ . So $a \in (3, \infty)$ and $γ = 3$ .

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

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JEE Main 2026 Apr 6 Shift 1, Mathematics Q3: Roots Lie In An Interval

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