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): You can put this solution on YOUR website!
 Step 1: Move all variables to left side and constants to right side



Use completing the square with both x and y expressions.

Step 2: factor out leading coefficients:



Step 3: Take 2nd coefficient, Halve it, then square it



Step 4: Multiply leading coefficient by number found in step 3 and add to right side.



Step 5: Factor expressions, notice they are perfect squares



Step 6: Divide equation by 15, In standard form the right side must be 1.



Now it is in standard form: A vertical Hyperbola with center (2,2)

 

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