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n200080 [17]
3 years ago
5

What are the vertex focus and directrix of the parabola with the given equation y=1/28(x-4)^2-5

Mathematics
2 answers:
patriot [66]3 years ago
8 0
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<span> = 4p(y - k)</span><span>, where the </span>focus<span> is (h, k+p) and the </span>directrix<span> is y = k - p. So, we need to determine the values from the equation.

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

focus= </span>(h, k+p) = 4,2
directrix = y = k - p = -12

Hope this answers the question.
Nata [24]3 years ago
8 0

The answer is:

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

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<u>Answer-</u>

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And

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According to the question,

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