Question

What is the focus, vertex, and directrix of the parabola described by `16x^2=y`?

Answer 1

Vertex is at `(0,0)`, directrix is `y= -1/64` and focus is at `(0,1/64)`.

Explanation:

`y=16x^2 or y = 16(x-0)^2+0` . Comparing with standard vertex form

of equation , `y= a(x-h)^2 +k ; (h,k)` being vertex , we find here

`h=0,k=0 , a =16`. So vertex is at `(0,0)` . Vertex is at

equidistance from focus and directrix situated at opposite sides.

since `a > 0` the parabola opens up . The distance of directrix from

vertex is `d=1/(4|a|) = 1/(4*16) =1/64` So directrix is `y= -1/64` .

Focus is at `0, (0+1/64) or (0,1/64)`.

graph{16x^2 [-10, 10, -5, 5]} [Ans]

Answer 2

`(0,1/64),(0,0),y=-1/64`

Explanation:

`"express the equation in standard form"`

`"that is "x^2=4py`

`rArrx^2=1/16y`

`"this is the standard form of a parabola with the y-axis"`
`"as its principal axis and vertex at the origin"`

`"if 4p is positive graph opens up, if 4p is"`
`"negative the graph opens down"`

`rArrcolor(blue)"vertex "=(0,0)`

`"by comparison " 4p=1/16rArrp=1/64`

`"focus "=(0,p)`

`rArrcolor(red)"focus "=(0,1/64)`

`"the directrix is a horizontal line below the origin"`

`"equation of directrix is " y=-p`

`rArrcolor(red)"equation of directrix " y=-1/64`