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JEE Main 7-April-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 7-April-2025 Paper - Shift 24 Marks
MCQ
Let $e_{1}$ and $e_{2}$ be the eccentricities of the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{25}=1$ and the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1$, respectively. If $\mathrm{b}<5$ and $\mathrm{e}_{1} \mathrm{e}_{2}=1$, 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 :
  • A
    $\frac{4}{5}$
  • B
    $\frac{3}{5}$
  • C
    $\frac{\sqrt{7}}{4}$
  • D
    $\frac{\sqrt{3}}{2}$
✓

Answer

B. $\frac{3}{5}$
$e_{1}^{2}=1-\frac{b^{2}}{25} \quad e_{2}^{2}=1-\frac{b^{2}}{16}$
$\therefore \mathrm{e}_{1}^{2} \mathrm{e}_{2}^{2}=1$
$\left(1-\frac{b^{2}}{25}\right)\left(1+\frac{b^{2}}{16}\right)=1$
$\Rightarrow 2+\frac{\mathrm{b}^{2}}{16}-\frac{\mathrm{b}^{2}}{25}-\frac{\mathrm{b}^{2}}{400}=1$
$\Rightarrow \frac{9 \mathrm{~b}^{2}}{400}=\frac{\mathrm{b}^{4}}{400}$
$b^{2}=9$
$\frac{x^{2}}{9}+\frac{y^{2}}{25}=1 \quad \frac{x^{2}}{16}-\frac{y^{2}}{9}=0$
$\mathrm{e}_{1} \sqrt{1-\frac{9}{25}} \quad \mathrm{e}_{2}=\frac{5}{4}$
$e_{1}=\frac{4}{5}$
Focii : $-(0, \pm 4) \quad( \pm 5,0)$
ellipse passing through all four foci
$\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$
$e=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$

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