Question 1138077
Find the vertex, focus, directrix, and focal width of the parabola. choices below  

{{{-(1/ 16)x^2 = y}}}

The standard form is {{{(x - h)^2 = 4p (y - k)}}}, where the focus is ({{{h}}},{{{ k + p}}}) and the directrix is {{{y = k - p}}}.

given: 
{{{-(1/ 16)x^2 = y}}}

{{{x^2 = (1/(1/-16))y}}}

{{{x^2 = -16y}}}

=>{{{h=0}}},{{{k=0}}}=>the vertex at:({{{0}}}, {{{0}}}) 

{{{ 4p=-16}}}=>{{{p=-4}}}

a focus at  ({{{h}}}, {{{k + p}}})=({{{0}}}, {{{-4}}}) 

a directrix at {{{y = k-p}}}=>{{{y=0 - (-4)=4}}}

 
The focal width of a parabola is the length of a segment that is parallel to the directrix and passing through the focus of a parabola. The length of this segment is {{{4p}}} units, or four time the length from the focus to the vertex.

focal width:{{{16}}}


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