JEE Main 2025 Apr 7 Shift 2, Mathematics Q18: Introduction to Ellipse

Since $b<5$, the ellipse $\dfrac{x^2}{b^2}+\dfrac{y^2}{25}=1$ has its major axis along $y$, with $e_1^2=1-\dfrac{b^2}{25}$. For the hyperbola, $e_2^2=1+\dfrac{b^2}{16}$.

The condition $e_1{e}_2=1$ gives $\left(1-\frac{b^2}{25}\right)\left(1+\frac{b^2}{16}\right)=1$. Expanding and simplifying leads to $b^2\left(\frac{1}{16}-\frac{1}{25}\right)=\frac{b^4}{400}\Rightarrow 9b^2=b^4\Rightarrow b^2=9$.

So the ellipse is $\frac{x^2}{9}+\frac{y^2}{25}=1$ with $e_1=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$, and the hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=1$ with $e_2=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$ (and indeed $e_1{e}_2=1$).

Ellipse foci (major axis along $y$, semi-major 5): $c=5\times \frac{4}{5}=4$, foci $(0,\pm 4)$.

Hyperbola foci (transverse axis along $x$, semi-transverse 4): $c=4\times \frac{5}{4}=5$, foci $(\pm 5,0)$.

An ellipse with axes along the coordinate axes through all four points $(0,\pm 4),(\pm 5,0)$ must be $\frac{x^2}{25}+\frac{y^2}{16}=1$ (major axis now along $x$).

Its eccentricity: $e=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$.