Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1. (2 points)

A. f(x) = − one twelfth (x − 5)2 + 2

B. f(x) = one twelfth (x − 5)2 + 2

C. f(x) = − one twelfth (x + 5)2 + 2

D. f(x) = one twelfth (x + 5)2 + 2

The focus is above the directrix, so the parabola opens upward—its vertical scale factor is positive (+1/12). The line of symmetry is x=-5, so the vertex form of the equation will have the factor (x -(-5))² = (x+5)². The choice that meets both these requirements is
D. f(x) = (1/12)(x + 5)² + 2

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Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1. (2 points) A. f(x) = − one twelfth (x − 5)2 + 2 B. f(x) = one twelfth (x − 5)2 + 2 C. f(x) = − one twelfth (x + 5)2 + 2 D. f(x) = one twelfth (x + 5)2 + 2

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Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1. (2 points)

A. f(x) = − one twelfth (x − 5)2 + 2

B. f(x) = one twelfth (x − 5)2 + 2

C. f(x) = − one twelfth (x + 5)2 + 2

D. f(x) = one twelfth (x + 5)2 + 2