Question 14 • Precalculus
For a conic of the form Ax² + Bxy + Cy² + … = 0, it is a hyperbola when B² − 4AC is positive. In choice C, A = 3, B = 1, and C = −9, so B² − 4AC = 1² − 4(3)(−9) = 109, which is positive. Therefore, choice C, 3x² + xy − 9y² + 2x + 5 = 0, represents a hyperbola.
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