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Lv 6
Find an equation for the hyperbola that satisfies the given conditions. Foci: (±12, 0), hyperbola passes through (16, 4)?
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- ?Lv 74 years ago
Foci: (±12,0)
Centre: (0,0)
c = Distance from centre to foci = 12
a² + b² = c² = 144
x²/a² − y²/b² = 1
Hyperbola passes through (16,4)
16²/a² − 4²/b² = 1
256b² − 16a² = a²b²
256b² − 16(144−b²) = (144−b²)b²
256b² − 2304 + 16b² = 144b² − b⁴
b⁴ + 128b² − 2304 = 0
(b² − 16) (b² + 144) = 0
b² = 16
a² = 144−16 = 128
Equation of hyperbola: x²/128 − y²/16 = 1
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