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What are the vertex focus and directrix of the parabola with the given equation

y=1/28(x-4)^2-5

2 Answers

2 votes

The answer is:

vertex: (4,-5); focus: (4,2); directrix: y=-12

User Tushu
by
7.8k points
2 votes
If the parabola is rotated so that its vertex is (h,k) and its axis of symmetry is parallel to the y-axis, it has an equation of (x - h)^2 = 4p(y - k), where the focus is (h, k+p) and the directrix is y = k - p. So, we need to determine the values from the equation.

y=1/28(x-4)^2-5
(x-4)^2 = 28(y+5)
(h,k) = (4, -5)
p = 7

focus=
(h, k+p) = 4,2
directrix = y = k - p = -12

Hope this answers the question.
User Szaman
by
7.2k points

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