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Hyperbola

Section: Mathematics

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Question : 2 of 98

Marks: +1, -0

Let

e_{1}

e_{2}

be two distinct roots of the equation

x^{2} - ax + 2 = 0

\{ a \in \mathbb{R}: e_{1}.

e_{2}

are the eccentricities of hyperbolas

\} = (\alpha, \beta)

\{ a \in \mathbb{R}: e_{1}.

e_{2}

are the eccentricities of an ellipse and a hyperbola, respectively

\} = (\gamma, \infty)

\alpha^{2} + \beta^{2} + \gamma^{2}

[JEE Main 6 Apr 2026 shift 1]

18
22
26
34

Solution:

e_{1} + e_{2} = a

e_{1} e_{2} = 2 \Rightarrow e_{2} = \frac{2}{e_{1}}

Case 1 : Both are eccentricities of hyperbola

Now,

e_{1} > 1 \text{ and } e_{2} > 1

\Rightarrow \frac{2}{e_{1}} > 1 \Rightarrow e_{1} < 2 \Rightarrow e_{1} \in (1,2)

Now,

a = e_{1} + \frac{1}{e_{1}}

Minimum occurs at

e_{1} = \sqrt{2}

\Rightarrow a_{\min} = 2 \sqrt{2}

if

e = 1,2

(end points)

a \longrightarrow 2,3 \;\; \Rightarrow a \in (2\sqrt{2}, 3)

\Rightarrow \alpha = 2 \sqrt{2} \text{ and } \beta = 3

Case 2 : One ellipse, one hyperbola

(e_{1}) \;\; (e_{2})

0 < e_{1} < 1 \text{ and } e_{2} < 1

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

Now

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

as

e_{1} \longrightarrow 1 \Rightarrow a \longrightarrow 3

e_{1} \longrightarrow 0 \Rightarrow a \longrightarrow \infty

\therefore (3, \infty) \Rightarrow r = 3

Now

\alpha^{2} + \beta^{2} + \gamma^{2} = 8 + 9 + 9 = 26

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