SOLUTION: Find the vertex, focus, directrix, and focal width of the parabola. Show the graph. {{{ (y-1)^2=16(x-2) }}}

Algebra ->  Rational-functions -> SOLUTION: Find the vertex, focus, directrix, and focal width of the parabola. Show the graph. {{{ (y-1)^2=16(x-2) }}}      Log On


   

Question 1175584: Find the vertex, focus, directrix, and focal width of the parabola. Show the graph.
+%28y-1%29%5E2=16%28x-2%29+
Answer by MathLover1(20850) About Me  (Show Source):
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Find the vertex, focus, directrix, and focal width of the parabola. Show the graph.
%28y-1%29%5E2=16%28x-2%29
%28y-k%29%5E2=4p%28x-h%29 is the standard equation for a right-left facing parabola with vertex at (h,+k )
and a focal length abs%28p%29
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%2Bp, 1 )
(2%2B4, 1 )
(6, 1+)