Question 442423
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Hi
Find the vertex and focus of the parabola.
y^2 = 36x – 216
Note: vertex form of a parabola, {{{x=a(y-k)^2 + h}}} where(h,k) is the vertex
  x = (1/36)y^2 + 6  V( 6,0) Parabola opening right a = 1/36 > 0
Note: standard form is {{{(y -k)^2 = 4p(x -h)}}}, where  the focus is (h + p, k)
 y^2 = 36(x-6)   4p = 36  p = 9
focus(6+9,0)or (15,0)
{{{drawing(300,300, -20,20,-20,20, grid(1),
circle(6,0,0.8),
circle(15, 0,0.8),
graph( 300, 300, -20,20,-20,20, sqrt(36x-216),-sqrt(36x-216)))}}}


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