SOLUTION: What type of conic section is the following equation? 4x2 + y2 = 36

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Question 812837: What type of conic section is the following equation?
4x2 + y2 = 36

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi
What type of conic section is the following equation?
4x2 + y2 = 36
x^2/9 + y^2/36 = 1 This is an Ellipse, see below.
See below descriptions of various conics
Standard Form of an Equation of a Circle is %28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2
where Pt(h,k) is the center and r is the radius
***Standard Form of an Equation of an Ellipse is %28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1+
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 ±sqrt%28a%5E2-b%5E2%29are the foci distances from center: a > b
Standard Form of an Equation of an Hyperbola opening up and down is:
%28y-k%29%5E2%2Fb%5E2+-+%28x-h%29%5E2%2Fa%5E2+=+1 with C(h,k) and vertices 'b' units up and down from center, 2b the length of the transverse axis
Foci sqrt%28a%5E2%2Bb%5E2%29units 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:
%28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 with C(h,k) and vertices 'a' units right and left of center, 2a the length of the transverse axis
Foci are sqrt%28a%5E2%2Bb%5E2%29 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), y=a%28x-h%29%5E2+%2Bk
where(h,k) is the vertex and x = h is the Line of Symmetry
The standard form is %28x+-h%29%5E2+=+4p%28y+-k%29, where the focus is (h,k + p)
the vertex form of a Parabola opening right(a>0) or left(a<0), x=a%28y-k%29%5E2+%2Bh
where(h,k) is the vertex and y = k is the Line of Symmetry