HyperbolahardPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Complex Equation Solvability: 9(α+β) = -10 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let e1e_1 and e2e_2 be two distinct roots of the equation x2−ax+2=0x^2-ax+2=0. Let the set {a∈R:e1,e2\{a\in\mathbb{R}:e_1,e_2 are eccentricities of hyperbolas}=(α,β)\}=(\alpha,\beta) and {a∈R:e1,e2\{a\in\mathbb{R}:e_1,e_2 are the eccentricities of an ellipse and a hyperbola respectively}=(γ,∞)\}=(\gamma,\infty). Then α2+β2+γ2\alpha^2+\beta^2+\gamma^2 is equal to
A1818
B2222
C2626correct
D3434
Solution
Step 1: For x2−ax+2=0x^2-ax+2=0: e1+e2=ae_1+e_2=a, e1e2=2e_1e_2=2.
⇒e2=2e1,a=e1+2e1.\Rightarrow e_2=\dfrac{2}{e_1},\qquad a=e_1+\dfrac{2}{e_1}.
 Step 2 — both hyperbolas (e1,e2>1e_1,e_2>1): e1>1e_1>1 and 2e1>1⇒e1<2\dfrac{2}{e_1}>1\Rightarrow e_1<2, so e1∈(1,2)e_1\in(1,2).
a′(e1)=1−2e12=0⇒e12=2⇒e1=2.a'(e_1)=1-\dfrac{2}{e_1^2}=0\Rightarrow e_1^2=2\Rightarrow e_1=\sqrt2.
amin⁡=2+22=2+2=22.a_{\min}=\sqrt2+\dfrac{2}{\sqrt2}=\sqrt2+\sqrt2=2\sqrt2.
 At e1→1+e_1\to1^+: a→1+2=3a\to1+2=3; at e1→2−e_1\to2^-: a→2+1=3a\to2+1=3. Since 22<32\sqrt2<3:
a∈(22,3)⇒α=22, β=3.a\in(2\sqrt2,3)\Rightarrow\alpha=2\sqrt2,\ \beta=3.
 Step 3 — e1e_1 ellipse, e2e_2 hyperbola (0<e1<10<e_1<1; then e2=2e1>2>1e_2=\dfrac{2}{e_1}>2>1): on (0,1)(0,1), a′=1−2e12<0a'=1-\dfrac{2}{e_1^2}<0. At e1→1−e_1\to1^-: a→3a\to3; at e1→0+e_1\to0^+: 2e1→∞⇒a→∞\dfrac{2}{e_1}\to\infty\Rightarrow a\to\infty.
⇒a∈(3,∞)⇒γ=3.\Rightarrow a\in(3,\infty)\Rightarrow\gamma=3.
 Step 4: ∴α2+β2+γ2=(22)2+32+32=8+9+9=26\therefore\alpha^2+\beta^2+\gamma^2=(2\sqrt2)^2+3^2+3^2=8+9+9=26. Correct answer: (3)
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