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Question 479772: Can you please put this equation into the standard form of a conic, step-by-step?
3y^2 + 20x = 23 + 5x^2 + 12y
Found 3 solutions by ewatrrr, lwsshak3, ccs2011:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi,
putting this equation into the standard form of a conic
3y^2 + 20x = 23 + 5x^2 + 12y
3y^2 - 12y - 5x^2 + 20x = 23
3(y-2)^2 -12 - 5(x-2)^2 +20 = 23
3(y-2)^2 -5(x-2)^2 = 15
 |Hyperbola opening up and down
Conics in General:
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 and a and b are the respective vertices distances from center.
Standard Form of an Equation of an Hyperbola opening right and left is:
 where Pt(h,k) is a center with vertices 'a' units right and left of center.
****Standard Form of an Equation of an Hyperbola opening up and down is:
 where Pt(h,k) is a center with vertices 'b' units up and down from center.
the vertex form of a parabola opening up or down,  where(h,k) is the vertex.
The standard form is  , where the focus is (h,k + p)
the vertex form of a parabola opening right or left,  where(h,k) is the vertex.
The standard form is  , where the focus is (h +p,k )
Answer by lwsshak3(11628)  (Show Source):
You can put this solution on YOUR website! Can you please put this equation into the standard form of a conic, step-by-step?
3y^2 + 20x = 23 + 5x^2 + 12y
**
3y^2 + 20x = 23 + 5x^2 + 12y
3y^2- 12y - 5x^2+ 20x = 23
complete the square
3(y^2-4y+4)-5(x^2-4x+4)=23+12-20=15
3(y-2)^2-5(x-2)^2=15
divide by 15
(y-2)^2/5-(x-2)^2/3=1
This is an equation of a hyperbola with vertical transverse axis of the standard form:
(y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the center of the hyperbola.
For given equation:
Center: (2,2)
a^2=5
a=√5
length of transverse axis=2a=2√5=4.47
b^2=3
b=√3
length of conjugate axis=2b=2√3=3.46
Answer by ccs2011(207)  (Show Source):
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