Let $\mathrm{e}_{1}$ and $\mathrm{e}_{2}$ 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 :

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Ellipse

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

b < 5

and

e_{1} 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) $\frac{3}{5}$ ✓ Correct
(C) $\frac{7}{4}$
(D) $\frac{3}{2}$

Answer: (B)

Solution

For the ellipse

\frac{x ^{2}}{b ^{2}} + \frac{y ^{2}}{25} = 1

with

b < 5

e_{1}^{2} = 1 - \frac{b ^{2}}{25}

; for the hyperbola

\frac{x ^{2}}{16} - \frac{y ^{2}}{b ^{2}} = 1

e_{2}^{2} = 1 + \frac{b ^{2}}{16}

. The condition

e_{1} e_{2} = 1

gives

(1 - \frac{b ^{2}}{25}) (1 + \frac{b ^{2}}{16}) = 1

, which simplifies to

9 b^{2} = b^{4}

, so

b^{2} = 9

. The ellipse then has foci

(0, \pm 4)

and the hyperbola

(\pm 5, 0)

. The ellipse through all four foci is

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

, with eccentricity

e = 1 - \frac{16}{25} = \frac{3}{5}

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Ellipse

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