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A hyperbola has foci (±5, 0) and transverse axis length 8. It is
Ax²/9 - y²/16 = 1
Bx²/16 + y²/9 = 1
Cx²/64 - y²/9 = 1
Dx²/16 - y²/9 = 1
Answer & Solution
Correct answer: D. x²/16 - y²/9 = 1
2a = 8 gives a = 4, so b² = c² - a² = 25 - 16 = 9 and the hyperbola is x²/16 - y²/9 = 1.
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