Let e₁ and e₂ 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 e₁e₂ = 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 :

<p>Let's find the eccentricities of the given ellipse and hyperbola, and then determine the eccentricity of an ellipse that passes through all four foci.</p> <p><strong>Step 1: Find $ e_1 $ for the Ellipse</strong></p> <p>The equation of the ellipse is:</p> <p>$ \frac{x^2}{b^2} + \frac{y^2}{25} = 1 $</p> <p>The eccentricity $ e_1 $ is given by:</p> <p>$ e_1^2 = 1 - \frac{b^2}{25} $</p> <p><strong>Step 2: Find $ e_2 $ for the Hyperbola</strong></p> <p>The equation of the hyperbola is:</p> <p>$ \frac{x^2}{16} - \frac{y^2}{b^2} = 1 $</p> <p>The eccentricity $ e_2 $ is given by:</p> <p>$ e_2^2 = 1 + \frac{b^2}{16} $</p> <p><strong>Step 3: Using the Product $ e_1 e_2 = 1 $</strong></p> <p>Given:</p> <p>$ e_1 e_2 = 1 $</p> <p>Thus:</p> <p>$ \left(1 - \frac{b^2}{25}\right)\left(1 + \frac{b^2}{16}\right) = 1 $</p> <p>Expanding gives:</p> <p>$ 1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1 $</p> <p>Simplifying:</p> <p>$ \frac{9b^2}{400} = \frac{b^4}{400} $</p> <p>Thus:</p> <p>$ b^2 = 9 $</p> <p><strong>Step 4: Determine Eccentricities $ e_1 $ and $ e_2 $</strong></p> <p>Substitute $ b^2 = 9 $:</p> <p>For the ellipse:</p> <p>$ e_1^2 = 1 - \frac{9}{25} = \frac{16}{25} $</p> <p>$ e_1 = \frac{4}{5} $</p> <p>For the hyperbola:</p> <p>$ e_2 = \frac{5}{4} $</p> <p><strong>Step 5: Find the Eccentricity of the New Ellipse</strong></p> <p>The new ellipse's equation is:</p> <p>$ \frac{x^2}{25} + \frac{y^2}{16} = 1 $</p> <p>The eccentricity $ e $ is:</p> <p>$ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} $</p> <p>Thus, the eccentricity of the ellipse that passes through all four foci is $ \frac{3}{5} $.</p>