Question 1068143
The equation of a hyperbola involves numbers that we call
{{{h}}} , {{{k}}} , {{{c}}} , {{{a}}} , and {{{b}}} .
We have to find those values.
We realize that vertices and foci are all on a line with {{{y=1}}} .
The hyperbola looks like this:  )  ( ,
and with the line {{{y=1}}} it looks like this: ----)--(---- .
All the important points of the hyperbola have the same y-coordinate,
and that makes all calculations easy.
That {{{y=1}}} line TRAVERSES all those points.
The segment of that line connecting the vertices
is called the traNsverse axis of the hyperbola
(watch for that N when you spell the word).
The center of a hyperbola (or of an ellipse)
is the midpoint of that segment connecting the vertices.
It is also the midpoint of the segment connecting the foci.
We call that point {{{"( h , k )"}}} .
In this case, it is easiest to see the center of this hyperbola
as the point halfway between foci (0,1) and (10,1) ,
point (5,1), with {{{system(x=5,y=1)}}} .
That calculation is easy mental math.
If someone wanted to see the calculation on paper,
the coordinates of that midpoint are the averages of the coordinates of the foci:
{{{h=(0+10)/2=10/2=5}}} and {{{k=(1+1)/2=2/2=1}}} .
So, we have {{{highlight(system(h=5,k=1))}}} .
The distance from the center of a hyperbola (or of an ellipse) to each focus
is called the focal distance, {{{c}}} .
In this case {{{highlight(c=5)=5-0}}} , the distance from center (5,1), to focus (0,1).
The distance from the center of a hyperbola (or of an ellipse) to each vertex
is called {{{a}}} .
In this case {{{highlight(a=4)=5-1}}} , the distance from center center (5,1), to vertex (1,1).
The number {{{b}}} (for a hyperbola, or for an ellipse)
is the distance from the center to points called co-vertices.
In a hyperbola, the numbers (or distances) {{{c}}} , {{{a}}} , and {{{b}}}
are lengths of sides of a right triangle,
related by {{{b^2+a^2=c^2}}} .
In this case, {{{b^2+4^2=5^2}}} ,
so {{{b^2+16=25}}} <--> {{{b^2+16=25}}} <--> {{{b^2=9}}} <--> {{{highlight(b=3)}}} .
Now, we can write the equation of the hyperbola as
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}} .
Substituting the number values we found for {{{h}}} , {{{k}}} , {{{c}}} , {{{a}}} , and {{{b}}} ,
the equation  is
{{{highlight((x-5)^2/4^2-(y-1)^2/3^2=1)}}} or {{{highlight((x-5)^2/16-(y-1)^2/9=1)}}} .
Some hyperbolas have equations of the form
{{{-(x-h)^2/b^2+(y-k)^2/a^2=1}}} or {{{(y-k)^2/a^2-(x-h)^2/b^2=1}}} ,
but when a coordinate value (like {{{y=1}}} ) is hared by vertices, foci, and center of a hyperbola,
the term with that coordinate gets the minus sign and the {{{b}}} .