JEE Main 2026 — 6 April, Morning Shift — Question 23

$e_{1}+e_{2}=a$ $e_{1} e_{2}=2 \Rightarrow e_{2}=\frac{2}{e_{1}}$ Case 1 : Both are eccentricities of hyperbola Now, $\mathrm{e}_{1}>1$ and $\mathrm{e}_{2}>1$ $\Rightarrow \frac{2}{\mathrm{e}_{1}}>1 \Rightarrow \mathrm{e}_{1}<2 \Rightarrow \mathrm{e}_{1} \in(1,2)$ Now, $\mathrm{a}=\mathrm{e}_{1}+\frac{1}{\mathrm{e}_{1}}$ Minimum occurs at $\mathrm{e}_{1}=\sqrt{2}$ $\Rightarrow \mathrm{a}_{\text {min }}=2 \sqrt{2}$ if $\mathrm{e}=1,2$ (end points) $\mathrm{a} \rightarrow 2,3 \Rightarrow \mathrm{a} \in(2 \sqrt{2}, 3)$ $\Rightarrow \alpha=2 \sqrt{2}$ and $\beta=3$ Case 2 : One ellipse, one hyperbola

$$\begin{aligned} & \left(\mathrm{e}_{1}\right) \\& 0<\mathrm{e}_{1}<1 \text { and } \mathrm{e}_{2}<1 \\& \mathrm{e}_{2}=\frac{2}{\mathrm{e}_{1}}>2 \end{aligned}$$

Now $\mathrm{a}=\mathrm{e}_{1}+\frac{2}{\mathrm{e}_{1}}$ as $\mathrm{e}_{1} \rightarrow 1 \Rightarrow \mathrm{a} \rightarrow 3$ $\mathrm{e}_{1} \rightarrow 0 \Rightarrow \mathrm{a} \rightarrow \infty$ $\therefore(3, \infty) \Rightarrow \mathrm{r}=3$ Now $\alpha^{2}+\beta^{2}+\gamma^{2}=8+9+9=26$