Question 1175584


Find the vertex, focus, directrix, and focal width of the parabola. Show the graph.

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

 {{{(y-k)^2=4p(x-h)}}} is the standard equation for a right-left facing parabola with vertex at ({{{h}}},{{{ k}}} )
and a focal length {{{abs(p)}}}

 so, 
the vertex is at: ({{{2}}},{{{1}}})

A parabola is the locus of points such that the distance to a point (the focus ) equals the distance to a line (the directrix).

{{{4p=16}}} => {{{p=4}}}
Parabola is symmetric around the x-axis and so the directrix is a line parallel to the y-axis, a distance  {{{-p}}} from the center  ({{{2}}},{{{1}}}) x-coordinate 

{{{x=2-p}}}
{{{x=2-4}}}
{{{x=-2 }}}

the focus lies a distance {{{p}}} from the center ({{{2}}},{{{1}}}) along the x-axis, and it will be at

({{{2+p}}}, {{{1}}} )
({{{2+4}}}, {{{1}}} )
({{{6}}}, {{{1 }}})


{{{drawing ( 600, 600, -10,10, -10, 10, 
circle(6,1,.12),locate(6,1,F(6,1)),
circle(2,1,.12),locate(2,1,C(2,1)),
green(line(-2,10,-2,-10)),
graph( 600, 600, -10,10, -10, 10, sqrt(16(x-2))+1, -sqrt(16(x-2))+1)) }}}