JEE Main 2025 (Online) 7th April Evening ShiftQuestion 1
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 $e_1e_2=1$, 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 :
Hyperbola
Solution
$e_1^2=1-\dfrac{b^2}{25}$, $e_2^2=1+\dfrac{b^2}{16}$; $e_1^2e_2^2=1\Rightarrow(25-b^2)(16+b^2)=400\Rightarrow b^4=9b^2$, $b^2=9$. Ellipse foci $(0,\pm4)$; hyperbola foci $(\pm5,0)$. The required ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ has $e^2=1-\dfrac{16}{25}=\dfrac{9}{25}$, so $e=\dfrac35$.