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Find the vertex, focus, directrix, and focal width of the parabola.

-1/40 x² = y


A) Vertex: (0, 0); Focus: (0, -10); Directrix: y = 10; Focal width: 160
B) Vertex: (0, 0); Focus: (-20, 0); Directrix: x = 10; Focal width: 160
C) Vertex: (0, 0); Focus: (0, -10); Directrix: y = 10; Focal width: 40
D) Vertex: (0, 0); Focus: (0, 10); Directrix: y = -10; Focal width: 10

User ConSod
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Final answer:

The vertex is at (0, 0), the focus is at (0, -10), the directrix is the line y = 10, and the focal width is 40. The correct option, according to these values, is option C.

Step-by-step explanation:

To find the vertex, focus, directrix, and focal width of the parabola described by the equation -1/40 x² = y, we first note that this is a vertical parabola since the x-term is squared and the coefficient is negative, indicating that it opens downwards.

The standard form of a parabola's equation that opens up or down is y-k=a(x-h)², where (h, k) is the vertex. In our equation, h = 0 and k = 0, so the vertex is at (0, 0).

The coefficient a is related to the focal length f, where a = -1/(4f). Hence, f = -10 since a = -1/40. The focus is a distance f from the vertex along the axis of symmetry of the parabola, so the focus is at (0, -10).

The directrix is a horizontal line f units opposite the focus from the vertex, giving us the equation y = 10. The focal width of a parabola is given by the absolute value of 4f, which is 4 * 10 = 40.

Finally, examining our calculations and the given options, we conclude that option C is the correct choice: Vertex: (0, 0); Focus: (0, -10); Directrix: y = 10; Focal width: 40.

User Brad Conyers
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