1. **State the problem:** We have an ellipse $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$ and a hyperbola $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$ with $b < 5$. Their eccentricities are $e_1$ and $e_2$ respectively, and satisfy $e_1 e_2 = 1$. We want the eccentricity of the ellipse passing through all four foci of these conics. 2. **Recall eccentricity formulas:** - Ellipse eccentricity: $e_1 = \frac{c_1}{a_1}$ where $a_1$ is semi-major axis and $c_1 = \sqrt{a_1^2 - b_1^2}$. - Hyperbola eccentricity: $e_2 = \frac{c_2}{a_2}$ where $a_2$ is semi-major axis and $c_2 = \sqrt{a_2^2 + b_2^2}$. 3. **Identify parameters for ellipse:** Given ellipse: $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$. Since $25 > b^2$ (because $b<5$), the major axis is along $y$-axis. So $a_1 = 5$, $b_1 = b$. Then $c_1 = \sqrt{a_1^2 - b_1^2} = \sqrt{25 - b^2}$. Eccentricity of ellipse: $$e_1 = \frac{c_1}{a_1} = \frac{\sqrt{25 - b^2}}{5}$$ 4. **Identify parameters for hyperbola:** Given hyperbola: $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$. Here $a_2 = 4$, $b_2 = b$. Foci distance: $$c_2 = \sqrt{a_2^2 + b_2^2} = \sqrt{16 + b^2}$$ Eccentricity of hyperbola: $$e_2 = \frac{c_2}{a_2} = \frac{\sqrt{16 + b^2}}{4}$$ 5. **Use the condition $e_1 e_2 = 1$:** $$\frac{\sqrt{25 - b^2}}{5} \times \frac{\sqrt{16 + b^2}}{4} = 1$$ Multiply both sides by 20: $$\sqrt{(25 - b^2)(16 + b^2)} = 20$$ Square both sides: $$(25 - b^2)(16 + b^2) = 400$$ Expand: $$400 + 25 b^2 - 16 b^2 - b^4 = 400$$ Simplify: $$400 + 9 b^2 - b^4 = 400$$ Subtract 400: $$9 b^2 - b^4 = 0$$ Rewrite: $$b^2 (9 - b^2) = 0$$ So either $b^2 = 0$ (not valid) or $b^2 = 9$. Since $b < 5$, $b = 3$. 6. **Find foci coordinates:** - Ellipse foci: $(0, \pm c_1) = (0, \pm \sqrt{25 - 9}) = (0, \pm 4)$. - Hyperbola foci: $(\pm c_2, 0) = (\pm \sqrt{16 + 9}, 0) = (\pm 5, 0)$. 7. **Ellipse passing through all four foci:** Equation: $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$. It passes through $(\pm 5, 0)$ and $(0, \pm 4)$. Substitute $(5,0)$: $$\frac{25}{A^2} + 0 = 1 \Rightarrow A^2 = 25$$ Substitute $(0,4)$: $$0 + \frac{16}{B^2} = 1 \Rightarrow B^2 = 16$$ 8. **Calculate eccentricity of this ellipse:** Major axis is along $x$ since $A=5 > B=4$. $$e = \frac{\sqrt{A^2 - B^2}}{A} = \frac{\sqrt{25 - 16}}{5} = \frac{3}{5} = 0.6$$ **Final answer:** The eccentricity of the ellipse passing through all four foci is $\boxed{0.6}$.