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

2 Answers
Binayaka C.
Jul 24, 2017

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]

Jim G.
Jul 24, 2017

#(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#

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