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AP EAMCET202124 Aug 2021Evening ShiftMathematicsHyperbolaActual

The equation of hyperbola whose eccentricity is 5 3 and distance between the foci is 10 units is

Options

  1. A16 x^2-9 y^2=16
  2. B16 x^2-9 y^2=9
  3. C16 x^2-9 y^2=-144
  4. D16 x^2-9 y^2=144

Correct answer

D. 16 x^2-9 y^2=144

Step-by-step solution

Given, e=5 / 3 and 2 a e=10 2 a ( 5 3 )=10 a=3 (a e)^2=a^2+b^2 (5)^2=3^2+b^2 b^2=25-9=16 Equation of hyperbola is x^2 9 - y^2 16 =1 16 x^2-9 y^2=144

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