Let e1 and e2 be two distinct roots of the equation x^2 - ax + 2 = 0. Let the sets {a ∈ R : e1 and e2 are the eccentricities of hyperbolas} = (α, β), and {a ∈ R : e1 and e2 are the eccentricities of an ellipse and a hyperbola, respectively} = (γ, ∞). Then a^2 + β^2 + γ^2 is equal to:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q3 – JEE-MAIN 2026

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Home Maths Conic SectionsQuestion #127

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, r e s p e c t i v e l y} = (γ, \infty)

. Then

a^{2} + β^{2} + γ^{2}

is equal to:

Clue 1 of 3: Properties of Roots and Eccentricity

Recall Vieta's formulas for the roots of a quadratic equation and the range of eccentricity for ellipses ( $0 < e < 1$ ) and hyperbolas ( $e > 1$ ).

🧠 Full Solution Path

● Video Walkthrough

Step 1: Analyze Quadratic Equation and Root Conditions✦ Active

Given the equation $x^{2} - a x + 2 = 0$ , let its distinct roots be $e_{1}$ and $e_{2}$ . From Vieta's formulas, we have $e_{1} + e_{2} = a$ and $e_{1} e_{2} = 2$ . Since $e_{1}$ and $e_{2}$ are eccentricities, they must be positive ( $e_{1} > 0, e_{2} > 0$ ). This implies $a = e_{1} + e_{2} > 0$ . For distinct real roots, the discriminant must be positive: $D = (- a)^{2} - 4 (1) (2) = a^{2} - 8 > 0$ , which means $a^{2} > 8$ . Combining $a > 0$ and $a^{2} > 8$ , we get $a > \sqrt{8} = 2 \sqrt{2}$ .

Step 2: Determine Ranges for 'a' based on Eccentricity Types○ Expand

Case 1: $e_{1}$ and $e_{2}$ are eccentricities of hyperbolas. For a hyperbola, $e > 1$ . Thus, $e_{1} > 1$ and $e_{2} > 1$ . This implies $(e_{1} - 1) (e_{2} - 1) > 0$ . Expanding, $e_{1} e_{2} - (e_{1} + e_{2}) + 1 > 0$ . Substituting Vieta's formulas, $2 - a + 1 > 0 ⟹ 3 - a > 0 ⟹ a < 3$ . Combining with $a > 2 \sqrt{2}$ , the range for $a$ is $(2 \sqrt{2}, 3)$ . This corresponds to $(α, β)$ , so $α = 2 \sqrt{2}$ and $β = 3$ .

Case 2: One root is an eccentricity of an ellipse ( $0 < e < 1$ ) and the other is an eccentricity of a hyperbola ( $e > 1$ ). This implies $(e_{1} - 1) (e_{2} - 1) < 0$ . Expanding, $e_{1} e_{2} - (e_{1} + e_{2}) + 1 < 0$ . Substituting Vieta's formulas, $2 - a + 1 < 0 ⟹ 3 - a < 0 ⟹ a > 3$ . Combining with $a > 2 \sqrt{2}$ , the range for $a$ is $(3, \infty)$ . This corresponds to $(γ, \infty)$ , so $γ = 3$ .

💡 Teacher's Secret Hint

Remember that $e_{1}$ and $e_{2}$ are interchangeable, so the order of ellipse/hyperbola doesn't change the condition $(e_{1} - 1) (e_{2} - 1) < 0$ .

Step 3: Calculate the Final Expression○ Expand

We need to calculate $α^{2} + β^{2} + γ^{2}$ . Using the values found:

α^{2} = (2 \sqrt{2})^{2} = 4 \times 2 = 8

β^{2} = 3^{2} = 9

γ^{2} = 3^{2} = 9

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

💡 Teacher's Secret Hint

Ensure to square each term correctly before summing them up.

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Maths Question 3 – JEE-MAIN 2026

Clue 1 of 3: Properties of Roots and Eccentricity

🧠 Full Solution Path