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Conic Sections
Hyperbola
Easy

Question

The foci of the ellipse x216+y2b2=1{{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1 and the hyperbola x2144y281=125{{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}} coincide. Then the value of b2{b^2} is :

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Solution

Key Concepts and Formulas

This problem requires a solid understanding of the standard forms and properties of ellipses and hyperbolas, particularly how to determine the location of their foci. The central idea is that if the foci of two conic sections coincide, they must be at the same coordinates, implying their focal distances from the center (origin in this case) are equal.

  1. For an Ellipse with the standard equation x2A2+y2B2=1{{{x^2}} \over {{A^2}}} + {{{y^2}} \over {{B^2}}} = 1:
    • If A2>B2A^2 > B^2 (major axis along the x-axis), the foci are at (±ce,0)(\pm c_e, 0), where $c_e^2

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