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e(1) is the eccentricity of the ellipse ...

`e_(1)` is the eccentricity of the ellipse `(x^(2))/(25)+(y^(2))/(b^(2))=1` and `e_(2)` is the eccentricity of the hyperbola `(x^(2))/(16)-(y^(2))/(b^(2))= 1` such that `e_(1),e_(2) = 1` find the value of b

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To solve the problem, we need to find the value of \( b \) given the eccentricities of an ellipse and a hyperbola. ### Step-by-Step Solution: 1. **Identify the equations and their parameters:** - The equation of the ellipse is given by: \[ \frac{x^2}{25} + \frac{y^2}{b^2} = 1 \] Here, \( a^2 = 25 \) and \( b^2 = b^2 \). - The equation of the hyperbola is given by: \[ \frac{x^2}{16} - \frac{y^2}{b^2} = 1 \] Here, \( a^2 = 16 \) and \( b^2 = b^2 \). 2. **Calculate the eccentricity of the ellipse (\( e_1 \)):** - The formula for the eccentricity of an ellipse is: \[ e_1 = \sqrt{1 - \frac{b^2}{a^2}} \] - Substituting \( a^2 = 25 \): \[ e_1 = \sqrt{1 - \frac{b^2}{25}} \] 3. **Calculate the eccentricity of the hyperbola (\( e_2 \)):** - The formula for the eccentricity of a hyperbola is: \[ e_2 = \sqrt{1 + \frac{b^2}{a^2}} \] - Substituting \( a^2 = 16 \): \[ e_2 = \sqrt{1 + \frac{b^2}{16}} \] 4. **Set up the equation using the given condition \( e_1 \cdot e_2 = 1 \):** - We have: \[ \sqrt{1 - \frac{b^2}{25}} \cdot \sqrt{1 + \frac{b^2}{16}} = 1 \] 5. **Square both sides to eliminate the square roots:** - This gives: \[ \left(1 - \frac{b^2}{25}\right) \left(1 + \frac{b^2}{16}\right) = 1 \] 6. **Expand the left-hand side:** - Expanding gives: \[ 1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1 \] - Simplifying this leads to: \[ \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 0 \] 7. **Combine the terms:** - To combine \( \frac{b^2}{16} \) and \( -\frac{b^2}{25} \), find a common denominator: \[ \frac{25b^2 - 16b^2}{400} - \frac{b^4}{400} = 0 \] - This simplifies to: \[ \frac{9b^2 - b^4}{400} = 0 \] 8. **Factor the equation:** - This can be factored as: \[ b^2(9 - b^2) = 0 \] - Thus, we have two solutions: \[ b^2 = 0 \quad \text{or} \quad b^2 = 9 \] 9. **Determine valid values:** - Since \( b^2 = 0 \) is not valid (as it would make the ellipse undefined), we take: \[ b^2 = 9 \] - Therefore, \( b = \sqrt{9} = 3 \). ### Final Answer: The value of \( b \) is \( 3 \).
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