SOLUTION: The equation is 9x^2+3y^2=27 What is the center, vertices, domain, range, and foci? Thank You so much!!

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Question 838344: The equation is 9x^2+3y^2=27
What is the center, vertices, domain, range, and foci?
Thank You so much!!

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi
BY DEFINITION (Need to know the Standard forms of the Conics: see below)
9x^2+3y^2=27 0r y^2/9 + x^2/3 = 1 is an ellipse C(0,0): major axis along y-axis
vertices: (sqrt(3),0)& (sqrt(3),0) AND (0,3)&(0,-3) foci (0,sqrt(6)) and (0,-sqrt(6))
| ±are the foci distances from center: a > b

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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