JEE Main 2025 — Hyperbola Question with Solution
JEE Main 2025 (7 Apr Shift 2)
Question
Let and be the eccentricities of the ellipse and the hyperbola , respectively. If and , then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
$\begin{aligned}
& \mathrm{e}_1^2=1-\frac{\mathrm{b}^2}{25} \quad \mathrm{e}_2^2=1-\frac{\mathrm{b}^2}{16} \\ & \therefore \mathrm{e}_1^2 \mathrm{e}_2^2=1 \\ & \left(1-\frac{\mathrm{b}^2}{25}\right)\left(1+\frac{\mathrm{b}^2}{16}\right)=1 \\ & \Rightarrow 2+\frac{\mathrm{b}^2}{16}-\frac{\mathrm{b}^2}{25}-\frac{\mathrm{b}^2}{400}=1 \\ & \Rightarrow \frac{9 \mathrm{~b}^2}{400}=\frac{\mathrm{b}^4}{400} \\ & \mathrm{~b}^2=9 \\ & \frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1 \quad \frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{9}=0 \\ & \mathrm{e}_1 \sqrt{1-\frac{9}{25}} \\ & \mathrm{e}_1=\frac{4}{5}
\end{aligned}(0, \pm 4) \quad( \pm 5,0)\begin{aligned}
& \frac{x^2}{25}+\frac{y^2}{16}=1 \\ & e=\sqrt{1-\frac{16}{25}}=\frac{3}{5}
\end{aligned}$
& \mathrm{e}_1^2=1-\frac{\mathrm{b}^2}{25} \quad \mathrm{e}_2^2=1-\frac{\mathrm{b}^2}{16} \\ & \therefore \mathrm{e}_1^2 \mathrm{e}_2^2=1 \\ & \left(1-\frac{\mathrm{b}^2}{25}\right)\left(1+\frac{\mathrm{b}^2}{16}\right)=1 \\ & \Rightarrow 2+\frac{\mathrm{b}^2}{16}-\frac{\mathrm{b}^2}{25}-\frac{\mathrm{b}^2}{400}=1 \\ & \Rightarrow \frac{9 \mathrm{~b}^2}{400}=\frac{\mathrm{b}^4}{400} \\ & \mathrm{~b}^2=9 \\ & \frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1 \quad \frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{9}=0 \\ & \mathrm{e}_1 \sqrt{1-\frac{9}{25}} \\ & \mathrm{e}_1=\frac{4}{5}
\end{aligned}(0, \pm 4) \quad( \pm 5,0)\begin{aligned}
& \frac{x^2}{25}+\frac{y^2}{16}=1 \\ & e=\sqrt{1-\frac{16}{25}}=\frac{3}{5}
\end{aligned}$
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This is a previous-year question from JEE Main 2025, covering the Hyperbola chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.