SOLUTION: where is the transverse axis for this hyperbola? 8y^2-5x^2=40

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Question 622715: where is the transverse axis for this hyperbola?
8y^2-5x^2=40

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,
transverse axis for hyperbola:
8y%5E2-5x%5E2=40
+y%5E2%2F5+-+x%5E2%2F8+=+1 C(0,0) Vertices (0,±sqrt%285%29)
Opens Up and Down along x = 0 or the y-axis. tranverse axis is the y-axis.
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 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 right and left is:
%28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 where Pt(h,k) is a center with vertices 'a' units right and left of center
and foci sqrt%28a%5E2%2Bb%5E2%29 units right and left of center along y = k
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 where Pt(h,k) is a center with vertices 'b' units up and down from center,
and foci sqrt%28a%5E2%2Bb%5E2%29units units up and down from center, along x = h
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.
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.
The standard form is %28y+-k%29%5E2+=+4p%28x+-h%29, where the focus is (h +p,k )