Question 971243
General Equation  of Hyperbola
with vertices (2,-1) and (2,-5) ; the Latus Rectum is 12
<pre>
We plot the two vertices and the center which is the midpoint
between them (2,-3), and sketch the hyperbola approximately, which must 
open up and down.

{{{drawing(400,400,-4,8,-8,4,
graph(400,400,-4,8,-8,4,-sqrt(x^2-4x+16)/sqrt(3)-3),
graph(400,400,-4,8,-8,4,sqrt(x^2-4x+16)/sqrt(3)-3),

circle(2,-1,0.15),circle(2,-1,0.13),circle(2,-1,0.11),circle(2,-1,0.09),circle(2,-1,0.07),circle(2,-1,0.05),circle(2,-1,0.03),circle(2,-1,0.01),

circle(2,-5,0.15),circle(2,-5,0.13),circle(2,-5,0.11),circle(2,-5,0.09),circle(2,-5,0.07),circle(2,-5,0.05),circle(2,-5,0.03),circle(2,-5,0.01),

circle(2,-3,0.15),circle(2,-3,0.13),circle(2,-3,0.11),circle(2,-3,0.09),circle(2,-3,0.07),circle(2,-3,0.05),circle(2,-3,0.03),circle(2,-3,0.01)




)}}}

Since the hyperbola opens upward and downward, it has the equation:

{{{(y-k)^2/a^2-(x-h)^2/b^2=1}}}

The center is (h,k) = (2,-3), the midpoint between vertices, and 
a = 2, the distance between the center and a vertex.

So we have everything but b

{{{(y-k)^2/a^2-(x-h)^2/b^2=1}}}

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

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

If you have studied the formula for the length of the
latus rectum which is {{{2b^2/a}}}, we can use that to 
find b:

{{{2b^2/a=12}}}
{{{2b^2/2=12}}}
{{{b^2=12}}}

we have the equation

{{{(y+3)^2/4-(x-2)^2/b^2=1}}}
{{{(y+3)^2/4-(x-2)^2/12=1}}}

That's the standard equation.  The general equation can be 
gotten by multiplying through by LCD=12, squaring the
binomials and simplifying: 

{{{x^2-3y^2-4x-18y-11=0}}}

There is another way to get this answer if you haven't
studied the latus rectum formula.  Let me know in the 
thank-you note form below if you haven't studied that 
formula, and I'll show you how to find b without using 
it.

Edwin</pre>