Question 968175
Consider the following equation and determine if it is a parabola, an ellipse, or a hyperbola and sketch it's graph:
4y^2+6x-2y+3x^2-15=2y^2+4x^2 
<pre>

{{{4y^2+6x-2y+3x^2-15=2y^2+4x^2}}}

Subtract the right side from both sides to get 0 on the right:

{{{2y^2+6x-2y-x^2-15=0}}}

We can tell this is the graph of a hyperbola because the x² and 
y² terms have opposite signs when all non-zero terms are one 
side and all like terms combined. 

Get the y terms together and the x terms together, and add 15
to both sides to get the contant term off the left side:

{{{2y^2-2y-x^2+6x=15}}}

Factor the coefficient of the y² term, which is 2, out of the 
first two terms on the left, skipping a space at the end of the
parentheses.

Factor the coefficient of the x² term, which is -1, out of the 
last two terms on the left, skipping a space at the end of the
parentheses.


{{{2(y^2-y+matrix(1,3,"","",""))-1(x^2-6x+matrix(1,3,"","",""))=15}}}  

Complete the square inside each parentheses:

In your head or on scratch paper,
1. Multiply the coefficient of y, which is -1, by 1/2, getting -1/2.
2. Square -1/2, getting (-1/2)² = +1/4
3. Add +1/4 in the space in the first parentheses, which amounts to
   adding 2 times +1/4 or +1/2 to the left side. So add +1/2 to the right 
   side.

1. Multiply the coefficient of x, which is -6, by 1/2, getting -3.
2. Square -3, getting (-3)² = +9
3. Add +9 in the space in the first parentheses, which amounts to
   adding -1 times +9 or -9 to the left side. So add -9 to the 
   right side.

{{{2(y^2-y+1/4)-1(x^2-6x+9)=15+1/2-9}}}

Factor each trinomial: {{{y^2-y+1/4=(y-1/2)(y-1/2)=(y-1/2)^2}}}
                       {{{x^2-6x+9=(x-3)(x-3)=(x-3)^2}}}
Combine the terms on the right {{{15+1/2-9=6+1/2=13/2}}}

{{{2(y-1/2)^2-1(x-3)^2=13/2}}}

Clear the fraction on the right by multiplying 
through by 2:

{{{4(y-1/2)^2-2(x-3)^2=13}}}

Get 1 on the right by dividing through by 13:

{{{4(y-1/2)^2/13-2(x-3)^2/13=1}}}

Divide numerator and denominator by the coefficients of
the squares of the binomials:

{{{expr(4/4)(y-1/2)^2/expr(13/4)}}}{{{""-""}}}{{{expr(2/2)(x-3)^2/expr(13/2)}}}{{{""=""}}}{{{1}}}

{{{(y-1/2)^2/expr(13/4)}}}{{{""-""}}}{{{(x-3)^2/expr(13/2)}}}{{{""=""}}}{{{1}}}

This is in standard form:

{{{(y-k)^2/a^2-(x-h)^2/b^2=1}}} which is the equation of a hyperbola which
opens like this: {{{drawing(20,25,8,12,8,12, graph(20,25,8,12,8,12,sqrt(1+(x-10)^2)+10),graph(20,25,8,12,8,12,-sqrt(1+(x-10)^2)+10) )}}}                                 
  
So {{{h=3}}}, {{{k=1/2}}}, {{{a^2=13/4}}}, {{{b^2=13/2}}}, {{{a=sqrt(13)/2}}}, {{{b=sqrt(13/2)=sqrt(expr(13/2)*expr(2/2))=sqrt(26)/2}}}

The center is (h,k) = {{{(matrix(1,3,3,",",1/2))}}}, the big green dot below.

We draw the defining rectangle with (h,k) as its center, its vertical
dimension is 2a = {{{sqrt(13)}}}, and its horizontal dimension is 2b = {{{sqrt(26)}}}.

We extend the diagonals of the defining rectangle, since they are the
asymptotes of the hyperbola. Then we sketch in the hyperbola:

{{{drawing(400,400,-3,10,-6.5,6.5,graph(400,400,-3,10,-6.5,6.5,(1+sqrt(2x^2-12x+31))/2),
graph(400,400,-3,10,-6.5,6.5,(1-sqrt(2x^2-12x+31))/2),
green(line(3-sqrt(13/2),1/2-sqrt(13/4),3+sqrt(13/2),1/2-sqrt(13/4)),

line(3+sqrt(13/2),1/2-sqrt(13/4),3+sqrt(13/2),1/2+sqrt(13/4)),

line(3-sqrt(13/2),1/2+sqrt(13/4),3+sqrt(13/2),1/2+sqrt(13/4)),

line(3-sqrt(13/2),1/2-sqrt(13/4),3-sqrt(13/2),1/2+sqrt(13/4)),

circle(3,0.5,0.15),circle(3,0.5,0.13),circle(3,0.5,0.11),circle(3,0.5,0.09),circle(3,0.5,0.07),circle(3,0.5,0.05),circle(3,0.5,0.03),circle(3,0.5,0.01),
line(-10,-8.692388155,10,5.449747468),line(-10,9.692388155,10,-4.449747468)




)  )}}}

Edwin</pre>