SOLUTION: What is the foci, vertices, center, and asymptotes for equation x=(1/2)(y+2)^2-(5/2)

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Question 822153: What is the foci, vertices, center, and asymptotes for equation x=(1/2)(y+2)^2-(5/2)
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
| V(-5/2, -2), y = -2, line of symmetry
(y+2)^2 = 2(x+5/2), 4p = 2, p = 1/2 |Foci: (-2,-2) & (-3,-2)
the vertex form of a Parabola opening right(a>0) or left(a<0),
where(h,k) is the vertex and y = k is the Line of Symmetry
The standard form is , where the focus is (h +p,k )
See below descriptions of various conics
Standard Form of an Equation of a Circle is
where Pt(h,k) is the center and r is the radius
Standard Form of an Equation of an Ellipse is
where Pt(h,k) is the center. (a variable positioned to correspond with major axis)
a and b are the respective vertices distances from center
and ±are the foci distances from center: a > b
Standard Form of an Equation of an Hyperbola opening up and down is:
with C(h,k) and vertices 'b' units up and down from center, 2b the length of the transverse axis
Foci units units up and down from center, along x = h
& Asymptotes Lines passing thru C(h,k), with slopes m = ± b/a
Standard Form of an Equation of an Hyperbola opening right and left is:
with C(h,k) and vertices 'a' units right and left of center, 2a the length of the transverse axis
Foci are units right and left of center along y = k
& Asymptotes Lines passing thru C(h,k), with slopes m = ± b/a

the vertex form of a Parabola opening up(a>0) or down(a<0),
where(h,k) is the vertex and x = h is the Line of Symmetry
The standard form is , where the focus is (h,k + p)
the vertex form of a Parabola opening right(a>0) or left(a<0),
where(h,k) is the vertex and y = k is the Line of Symmetry
The standard form is , where the focus is (h +p,k )

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