SOLUTION: Identify the coordinates of the vertex, and focus, the equations of the axis of symmetry and directirix, and the directions of opening of the parabola with the given equation. Then

Question 458250: Identify the coordinates of the vertex, and focus, the equations of the axis of symmetry and directirix, and the directions of opening of the parabola with the given equation. Then find the length of the latus rectum and sketch a graph.
x=16y^2
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Identify the coordinates of the vertex, and focus, the equations of the axis of symmetry and directirix, and the directions of opening of the parabola with the given equation. Then find the length of the latus rectum and sketch a graph.
x=16y^2.
..
Standard form for parabola: (y-k)^2=4p(x-h), (h,k) being the (x,y) coordinates of the vertex.
Rewrite given parabola
y^2=(1/16)x
vertex (0,0)
4p=1/16
p=1/64
axis of symmetry y=0 or x-axis
parabola opens rightwards
focus (0,1/64)
directrix x=-1/64
latus rectum or focal width=4p=1/16
see graph below: (sorry, I don't have the means to provide more detail on the graph, but you should have enough information to fill it in.)
..

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