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Anonymous asked in Education & ReferenceHomework Help · 1 month ago

# Someone help me with my hw  PLEASSSEE?

find the vertex, focus and directrix of y^2+4y+12x+12=0

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• y^2+4y+12x+12=0

this is a parabola opening to the left.  To find the vertex, focus and directrix, you need to rewrite it in the form:

x = 4p(y - k)^2 + h

where (h,k) is the vertex and p is the distance from the vertex to the focus (and vertex to directrix.

now, to get it into form requires completing the square:

12x = -(y^2 + 4y) - 12

x = (-1/12)(y2 + 4y + 4) - 12 + 1/3

x = (-1/12)(y + 2)^2 - 35/3

vertex:  (-35/3, -2)

since 4p = -1/12

p = -1/48

focus: [(-35/3 - 1/48),-2] = (-11.6865, -2)

directrix:  x = -2 + 1/48 = -1.9792

hope this helps!

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•  y^2 + 4y + 12x + 12 = 0

parabola

focus | (-11/3, -2)≈(-3.66667, -2)

vertex | (-2/3, -2)≈(-0.666667, -2)

semi-axis length | 3

focal parameter | 6

eccentricity | 1

directrix | x = 7/3

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