Find the vertex, focus, directrix, and focal width of the parabola.
-1/16x^2 = y

a. vertex: (0, 0); focus: (0, -4); directrix: y = 4; focal width: 16
b. vertex: (0, 0); focus: (-8, 0); directrix: x = 4; focal width: 64
c. vertex: (0, 0); focus: (0, 4); directrix: y = -4; focal width: 4
d. vertex: (0, 0); focus: (0, -4); directrix: y = 4; focal width: 64

Vertex: (0, 0); Focus: (0, -4); Directrix: y = 4; Focal width: 16 ⇒ answer (a)

Step-by-step explanation:

* Lets revise some facts about the parabola

- Standard form equation for a parabola of vertex at (0 , 0)

- If the equation is in the form  x² =  4py, then  

- The axis of symmetry is the y-axis,  x = 0  

- 4p  equal to the coefficient of y in the given equation to

  solve for  p

- If  p  >  0, the parabola opens up.

- If  p <  0, the parabola opens down.

- Use  p  to find the coordinates of the focus,  (0  ,  p)

- Use  p  to find equation of the directri ,   y= − p

- Use  p  to find the endpoints of the focal diameter,  (±2p  ,  p)

* Now lets solve the problem

- The vertex of the parabola is (0 , 0)

∵ -1/16x² = y ⇒ multiply each side by -16

∴ x² = -16y

∴ 4p = -16 ⇒ ÷ 4 tbe both sides

∴ p = -4

∵ The focus is (0 , p)

∴ The focus is (0 , -4)

∵ The directrix is y = -p

∴ The directrix is y = -(-4) = 4 ⇒ y = 4

∵ The endpoints of the focal diameter,  (±2p  ,  p)

∴ The focal width = 2p - (-2p) = 4p

∴ The focal width = 4 ×  I-4I = 16

* Vertex: (0, 0); Focus: (0, -4); Directrix: y = 4; Focal width: 16

step-by-step explanation: