SOLUTION: What is the solution of hyperbola 2x^2+2y^2-8x+12y+2=0?

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Question 650104: What is the solution of hyperbola 2x^2+2y^2-8x+12y+2=0?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,
2x^2+2y^2-8x+12y+2=0
2(x-2)^2 + 2(y+3)^2 = -2 + 8 + 18
%28x-2%29%5E2+%2B+%28y%2B3%29%5E2+=+12
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)
and length of the latus rectum is 4p.
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
The standard form is %28y+-k%29%5E2+=+4p%28x+-h%29, where the focus is (h +p,k )
and length of the latus rectum is 4p.