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Find the equation of hyperbola whose foci are (±5, 0) and the conjugate axis is of the length 8. Also, find its eccentricity. |
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Answer» Foci of the equation is in the form (±c, 0) Equation of the hyperbola will be of the form: x2/a2 – y2/b2 = 1 Whose foci at (±c, 0) and conjugate axis is 2b On comparing, we get 2b = 8 ⇒ b = 4 And c = 5 Again, We know, c2 = a2 + b2 25 = 16 + a2 or a2 = 9 Equation (1)⇒ x2/9 – y2/16 = 1 Which is required equation. Again, Eccentricity (e) = c/a = 5/3 |
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