SOLUTION: 4x^2-3y^2+16x+2y=25/3 Identify if it is a hyperbola, parabola, or ellipse and label the center point, vertex points, focus points, asymptote equations, a length, b length, and c

The equation is 

The center is 

The vertices are the points  

The foci are the points 

The equations of the asymbtotes in point slope form are



How'd I get all that?

1. I knew it was a hyperbola because the x² and y² terms in

 have coefficients of opposite
signs when on the same side of the equation.  But I didn't
know whether it was going to be a hyperbola 
like this:  or like this: 

So I started with


 
Get the  term next to the x² term and the y term next
to the y² term



On the left, factor out 4 out of the first two terms 
and -3 out of the last two terms.  The unusual part
is having to factor -3 out of a +2y, for you never had
to do such a thing in basic algebra.  Just remembr that
"to factor out" means "to divide by", so you divide +2y
by -3 and you get : 



Then inside those parentheses you must complete the square:

In the first parentheses:

1. Multiply the coefficient of the x term, which is +4, by ,
   getting +2.
2. Square +2 getting +4.
3. Put +4-4 after the x term. (That is, add it and subtract 4, which
   amounts to adding 0)

 

In the second parentheses:

1. Multiply the coefficient of the y term, which is , by ,
   getting .
2. Square  getting .
3. Put  after the y term. (That is, add it and subtract , which
   amounts to adding 0)



Group the first three terms in each parentheses:



Factor  as  and then as 

Factor  as  and then as  

and so we have:



Now we have to remove the BIG parentheses by distributing 
without disturbing the smaller parentheses, like this:



Subtract the  from both sides since that will make a whole
number on the right:









Now we get 1 on the right by dividing through by 24





Now that's in the form of a hyperbola like this .  



, , , 

So we have the center.

The foci ar points which are "a" units right and left of the center.

The defining rectangle is 2a units wide and 2b units high.

It's extended diagonals are the asymptotes.

To find the foci we calculate c from c² = a²+b², and they are c
units right and left of the center.

The asymptotes have slopes  and pass through the 
center, so you can find them from the point-slope form.


Edwin