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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Hyperbola

Let e_1 and e_2 be the eccentricities of the ellipse x^2 ~b^2 + y^2...

Let e1e_1 and e2e_2 be the eccentricities of the ellipse x2 b2+y225=1\frac{\mathrm{x}^2}{\mathrm{~b}^2}+\frac{\mathrm{y}^2}{25}=1 \quad and the hyperbola x216−y2 b2=1\quad \frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, respectively. If b<5\mathrm{b} \lt 5 and e1e2=1\mathrm{e}_1 \mathrm{e}_2=1, 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 :

Options

  1. A45\frac{4}{5}
  2. B35\frac{3}{5}
  3. C74\frac{\sqrt{7}}{4}
  4. D32\frac{\sqrt{3}}{2}

Correct answer

B. 35\frac{3}{5}

Step-by-step solution

<br/>e12=1−b225e22=1−b216∴e12e22=1(1−b225)(1+b216)=1⇒2+b216−b225−b2400=1⇒9 b2400=b4400 b2=9x29+y225=1x216−y29=0e11−925e1=45<br/>\begin{aligned}<br/>& \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}<br/>\end{aligned}
Focii : - (0,±4)(±5,0)(0, \pm 4) \quad( \pm 5,0)
ellipse passing through all four foci
<br/>x225+y216=1e=1−1625=35<br/>\begin{aligned}<br/>& \frac{x^2}{25}+\frac{y^2}{16}=1 \\ & e=\sqrt{1-\frac{16}{25}}=\frac{3}{5}<br/>\end{aligned}
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