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Fine the vertex, focus and directrix of Y+7=1/12(x-6)^2

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Answer to a math question Fine the vertex, focus and directrix of Y+7=1/12(x-6)^2

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Fred

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1. The given equation is in the form

y + 7 = \frac{1}{12}(x - 6)^2

.

2. Convert it to the standard form of a parabola

y - k = \frac{1}{4p}(x - h)^2

, where

(h, k)

is the vertex.

- Here,

k = -7

h = 6

, and

\frac{1}{4p} = \frac{1}{12}

.

3. Calculate [Step-by-Step]

p = 3

.

Vertex:

(6, -7)

Focus: The focus is at

(h, k + p)

(6,-7+3)=(6,-4)

Directrix: The directrix is at

y = k - p

y=-7-3=-10

Answer:

Vertex:

(6, -7)

Focus:

(6,-4)

Directrix:

y=-10

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Fine the vertex, focus and directrix of Y+7=1/12(x-6)^2

Answer to a math question Fine the vertex, focus and directrix of Y+7=1/12(x-6)^2

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