Let e₁ and e₂ be the eccentricities of the ellipse x²/b² + y²/25 = 1 and the hyperbola x²/16 - y²/b² = 1, respectively. If b is less than 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:
❓ Question Let e 1 e_{1} and e 2 e_{2} be the eccentricities of Ellipse: x 2 b 2 + y 2 25 = 1 and Hyperbola: x 2 16 − y 2 b 2 = 1 respectively. If b < 5 b<5 and e 1 e 2 = 1 e_{1}e_{2}=1 , then find the eccentricity of the ellipse whose axes are along the coordinate axes and which passes through all four foci (two of the given ellipse and two of the hyperbola). 🖼️ Question Image ✍️ Short Solution Step 1 — Find e 1 e_1 e 1 and e 2 e_2 e 2 relations For ellipse: x 2 b 2 + y 2 25 = 1 \dfrac{x^2}{b^2} + \dfrac{y^2}{25} = 1 Major axis = along y y y -axis (since 25 > b 2 25 > b^2 ). Eccentricity: e 1 = 1 − b 2 25 . e_1 = \sqrt{1 - \frac{b^2}{25}}. . For hyperbola: x 2 16 − y 2 b 2 = 1 Eccentricity: e 2 = 1 + b 2 16 . Given e 1 e 2 = 1 e_1 e_2 = 1 : 1 − b 2 25 1 + b 2 16 = 1. Step 2 — Solve for b b b Square both sides: ( 1 − b 2 25 ) ( 1 + b 2 16 ) = 1. Expand: 1 + b 2 16 − b 2 25 − b 4 400 = 1. Simplify: b 2 16 − b ...