Question

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

-1/40x^2=y

Answer 1

answers:
Vertex: (0, 0); Focus: (0, -10); Directrix: y = 10; Focal width: 160

Vertex: (0, 0); Focus: (-20, 0); Directrix: x = 10; Focal width: 160

Vertex: (0, 0); Focus: (0, -10); Directrix: y = 10; Focal width: 40

Vertex: (0, 0); Focus: (0, 10); Directrix: y = -10; Focal width: 10

Answer 2

well write in the form

\[x^2 = 4ay\]

so your question becomes

\[x^2 = -40y\]

so the vertex is at the origin and the parabola is concave down.

\[x^2 = - 4 \times 10y\]

the value of a, the focal length is 10

and will be above the focus..

so the directrix is y = 10

focus is 10 below the vertex and on the line of symmetry x = 0

(0, -10)

so all you need to do is find the width of the parabola...

Answer 3

so the focal width is 40?

Answer 4

@campbell_st