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Write the equation for a parabola with the given characteristics: Focus (-5, -2); opens up; contains (-13, -2).

a. (x-5)² = -16(y-6)
b. (x+8)² = 1/6(y-3)
c. (x-8)² = -6(y+3)
d. (x+5)² = 16(y+6)

User Keynan
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1 Answer

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Final answer:

None of the provided options for the equation of a parabola match the given characteristics of having a focus at (-5, -2), opening up, and containing the point (-13, -2), because the parabola's vertex also lies on the horizontal line y = -2, which invalidates the possibility of the parabola having a positive 'p' value as needed for an upward opening parabola with a focus below the vertex.

Step-by-step explanation:

The equation for a parabola that has a focus at (-5, -2), opens up, and contains the point (-13, -2) can be determined based on the standard form of a parabola opening upwards, which is (x - h)² = 4p(y - k), where (h, k) is the vertex of the parabola and p is the distance from the vertex to the focus. Since the parabola contains the point (-13, -2) and this point lies on the same horizontal line as the focus, it indicates the vertex also lies on this line. Thus, the vertex has coordinates (-5, k) and the parabola's equation will be of the form (x + 5)² = 4p(y - k). Given the parabola opens upward and the focus is below the vertex, p is positive. If we assume the distance between the vertex and the focus is p, then p = 0 since they have the same y-coordinate, which is not possible. Hence, no equation provided exactly matches these characteristics.

User J Freebird
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