Question 854012: Identify the vertex, focus and directrix of the parabola with the equation x^2-6x-8y+49=0
Found 2 solutions by ewatrrr, lwsshak3:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi,
x^2-6x-8y+49=0
8y = (x-3)^2 + 40
y = (1/8)(x-3)^2 + 5 Parabola opening right, y = 5, line of symmetry
V(3,5),  ,  , Focus(5,5), Directrix x = 1
Answer by lwsshak3(11628)  (Show Source):
You can put this solution on YOUR website! Identify the vertex, focus and directrix of the parabola with the equation x^2-6x-8y+49=0
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x^2-6x-8y+49=0
x^2-6x-8y=-49
complete the square:
(x^2-6x+9)-8y=-49+9
(x-3)^2=8y-40
(x-3)^2=8(y-5)
This is an equation of a parabola that opens up:
Its basic equation: (x-h)^2=4p(y-k), (h,k)=coordinates of vertex
For given problem:
vertex:(3,5)
axis of symmetry: x=3
4p=8
p=2
focus: (3,7)(p-distance above vertex on the axis of symmetry)
directrix: y=3(p-distance below vertex on the axis of symmetry)
see graph below as a visual check:
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