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

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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 :

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

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