Question 1193003
<pre>
Instead of doing your homework for you, I'll change 9 to 16
and do this one instead: 

{{{y^2 -16x^2 =-16}}}

You will do yours exactly like this, step by step:

The standard form of a hyperbola is either

{{{(x-h)^2/a^2+(y-k)^2/b^2}}}{{{""=""}}}{{{1}}} for a hyperbopla like this " )( "

or  

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

Either way, we get 1 on the right side, by dividing through by -16

{{{y^2/(-16) -16x^2/(-16) =(-16)/(-16)}}}

Both terms on the left must have a denominator showing, so we put
1 under the second term on the left:

{{{-y^2/16+x^2/1 =1}}}

Reverse the terms on the left, so that the "plus" term is first:

{{{x^2/1-y^2/16=1}}}

We finish by rewriting x as (x-0) and y as (y-0)

{{{(x-0)^2/1-(y-0)^2/16=1}}}

Now that we have it in standard form, we see that it is a hyperbola
that looks like this " )( ".

Comparing it to {{{(x-h)^2/a^2+(y-k)^2}}}{{{""=""}}}{{{1}}}

center = (h,k) = (0,0), a<sup>2</sup> = 1, b<sup>2</sup> = 16, so
a = semi-transverse axis = √1 = 1 and 
b =semi-conjugate axis = √16 = 4

The defining rectangle has corners (h±a,k±b) = (0±1,0±4) or
(1,4), (-1,4), (1,-4), (-1,-4), and the asymptotes are the
extended diagonals of the defining rectangle:

{{{drawing(200,400,-5,5,-10,10,

graph(200,400,-5,5,-10,10),
green(line(-1,-4,-1,4),line(-1,4,1,4),line(1,4,1,-4),line(1,-4,-1,-4)) )}}} {{{drawing(200,400,-5,5,-10,10,
green(line(12,-48,-12,48),line(-12,-48,12,48)),
graph(200,400,-5,5,-10,10),
green(line(-1,-4,-1,4),line(-1,4,1,4),line(1,4,1,-4),line(1,-4,-1,-4)) )}}} {{{drawing(200,400,-5,5,-10,10,

graph(200,400,-5,5,-10,10),
green(line(-1,-4,-1,3),line(-1,4,1,4),line(1,4,1,-4),line(1,-4,-1,-4)) )}}} {{{drawing(200,400,-5,5,-10,10,
green(line(12,-48,-12,48),line(-12,-48,12,48)),
graph(200,400,-5,5,-10,10,4sqrt(x^2-1)),
graph(200,400,-5,5,-10,10,-4sqrt(x^2-1)),
green(line(-1,-3,-1,3),line(-1,3,1,3),line(1,3,1,-3),line(1,-3,-1,-3)) )}}} 

The vertices are the ends of the transverse axis (1,0) and (-1,0),

the foci are just beyond the vertices.

To find the foci, we must find c by the Pythagorean relation for
hyperbolas:

{{{c^2=a^2+b^2}}}
{{{c^2=1^2+4^2}}}
{{{c^2=1+16}}}
{{{c^2=17}}}
{{{c=sqrt(17)}}}

So the foci are {{{(matrix(1,3,1,",", "" +- sqrt(17)))}}}

I'll draw them in:
{{{drawing(200,400,-6,6,-10,10,
circle(sqrt(17),0,.1),circle(sqrt(17),0,.15),circle(sqrt(17),0,.125),circle(sqrt(17),0,.2),circle(sqrt(17),0,.175),circle(sqrt(17),0,.23),

circle(-sqrt(17),0,.1),circle(-sqrt(17),0,.15),circle(-sqrt(17),0,.125),circle(-sqrt(17),0,.2),circle(-sqrt(17),0,.175),circle(-sqrt(17),0,.23),

green(line(12,-48,-12,48),line(-12,-48,12,48)),
graph(200,400,-6,6,-10,10,4sqrt(x^2-1)),
graph(200,400,-6,6,-10,10,-4sqrt(x^2-1)),
green(line(-1,-3,-1,3),line(-1,3,1,3),line(1,3,1,-3),line(1,-3,-1,-3)) )}}}

We find the equation of the asymptote that leans to the right.
It passes through the points (0,0) and (1,4).  It has slope
{{{m=(y[2]-y[1])/(x[2]-x[1]) = (4-0)/(1-0) = 4/1 = 4}}}.

It has equation
{{{y-y[1]=m(x-x[1])}}}
{{{y-0=4(x-0)}}}
{{{y=4x}}}

We find the equation of the asymptote that leans to the left.
It passes through the points (0,0) and (-1,4).  It has slope
{{{m=(y[2]-y[1])/(x[2]-x[1]) = (4-0)/(-1-0) = 4/(-1) = -4}}} 

It has equation
{{{y-y[1]=m(x-x[1])}}}
{{{y-0=-4(x-0)}}}
{{{y=-4x}}}

Now do yours the same way, step-by-step.

Edwin</pre>