Question 196562
find the center, foci, the length of one of the two axes (transverse or conjugate) which is parallel to the y-axis, and the two asymptotes.

{{{(1/10)(x-1)^2 - (y-1)^2= 1}}}

<pre><font size = 4 color = "indigo"><b>

Rewrite that as:

{{{((x-1)^2)/10 - ((y-1)^2)/1 = 1}}}

and compare with

{{{((x-h)^2)/a^2 - ((y-k)^2)/b^2 = 1}}}
 
{{{h = 1}}}, {{{k=1}}}, 
 
{{{a^2=10}}}, so {{{a=sqrt(10)}}}
 
{{{b^2=1}}}, so {{{b=1}}}
 
The center (h,k) = (1,1)
 
We start out plotting the center C(h,k) = C(1,1)
 
{{{drawing(400,400,-6,8,-6,8,
line(.9,1,1.1,1), line(1,.9,1,1.1), locate(1.2,1.5,"C(1,1)"),
graph(400,400,-6,8,-6,8) )}}}
 
Next we draw the left semi-transverse axis,
which is a segment {{{a=sqrt(10)}}} units long horizontally 
left from the center.  This semi-transverse 
axis ends up at one of the two vertices ({{{1-sqrt(10)}}},1).
                    
We'll call it V1({{{1-sqrt(10)}}},1).:
 
{{{drawing(400,400,-6,8,-6,8,
graph(400,400,-6,8,-6,8),

line(.9,1,1.1,1), line(1,.9,1,1.1), locate(1.2,1.5,"C(1,1)"),
line(1-sqrt(10),1,1,1), locate(-5.7,1.5,V1(1-sqrt(10),1))
 )}}}
 
Next we draw the right semi-transverse axis,
which is a segment {{{a=sqrt(10)}}} units long horizontally 
right from the center. This other semi-transverse
axis ends up at the other vertex ({{{1+sqrt(10)}}},1).
We'll call it V2({{{1+sqrt(10)}}},1).:
 
{{{drawing(400,400,-6,8,-6,8,
 
line(.9,1,1.1,1), line(1,.9,1,1.1), locate(1.2,1.5,"C(1,1)"),
line(1-sqrt(10),1,1,1), locate(-5.7,1.6,V1(1-sqrt(10),1)),
line(1+sqrt(10),1,1,1), locate(4.5,1.6,V2(1+sqrt(10),1)),
graph(400,400,-6,8,-6,8)
  )}}}
 
That's the whole transverse ("trans"="across",
"verse"="vertices", the line going across from
one vertex to the other. It is 2a in length,
so the length of the transverse axis is {{{2a=2sqrt(10)}}}
 
Next we draw the upper semi-conjugate axis,
which is a segment b=1 units long verically 
upward from the center.  This semi-conjugate
axis ends up at (1,2).
 
{{{drawing(400,400,-6,8,-6,8,
 
line(.9,1,1.1,1), line(1,.9,1,1.1), locate(1.2,1.5,"C(1,1)"),
line(1-sqrt(10),1,1,1), locate(-5.7,1.6,V1(1-sqrt(10),1)),
line(1+sqrt(10),1,1,1), locate(4.5,1.6,V2(1+sqrt(10),1)),
line(1,1,1,2), locate(.8,2.7,"(1,2)"),

 graph(400,400,-6,8,-6,8)
  )}}}
 
Next we draw the lower semi-conjugate axis,
which is a segment b=1 units long verically 
downward from the center.  This semi-conjugate
axis ends up at (1,0). 

{{{drawing(400,400,-6,8,-6,8,
 
line(.9,1,1.1,1), line(1,.9,1,1.1), locate(1.2,1.5,"C(1,1)"),
line(1-sqrt(10),1,1,1), locate(-5.7,1.6,V1(1-sqrt(10),1)),
line(1+sqrt(10),1,1,1), locate(4.5,1.6,V2(1+sqrt(10),1)),
line(1,1,1,2), locate(.8,2.7,"(1,2)"), 
line(1,1,1,0), locate(.8,-.8,"(1,0)"),
graph(400,400,-6,8,-6,8)
  )}}}
 
That's the complete conjugate axis. It is 2b in length,
so the length of the transverse axis is 2b=2(1)=2
 
Next we draw the defining rectangle which has the
ends of the transverse and conjugate axes as midpoints
of its sides:
 
{{{drawing(400,400,-6,8,-6,8,
 
line(.9,1,1.1,1), line(1,.9,1,1.1), locate(1.2,1.5,"C(1,1)"),
line(1-sqrt(10),1,1,1), locate(-5.7,1.6,V1(1-sqrt(10),1)),
line(1+sqrt(10),1,1,1), locate(4.5,1.6,V2(1+sqrt(10),1)),
line(1,1,1,2), locate(.8,2.7,"(1,2)"), 
line(1,1,1,0), locate(.8,-.8,"(1,0)"),
graph(400,400,-6,8,-6,8), rectangle(1-sqrt(10),0,1+sqrt(10),2)
  )}}}


Next we draw and extend the two diagonals of this defining
rectangle:


{{{drawing(400,400,-6,8,-6,8,
 
line(.9,1,1.1,1), line(1,.9,1,1.1), locate(1.2,1.5,"C(1,1)"),
line(1-sqrt(10),1,1,1), locate(-5.7,1.6,V1(1-sqrt(10),1)),
line(1+sqrt(10),1,1,1), locate(4.5,1.6,V2(1+sqrt(10),1)),
line(1,1,1,2), locate(.8,2.7,"(1,2)"), 
line(1,1,1,0), locate(.8,-.8,"(1,0)"),
graph(400,400,-6,8,-6,8), rectangle(1-sqrt(10),0,1+sqrt(10),2),
line(-7,1-8/sqrt(10),9,1+8/sqrt(10)),
line(-7,1+8/sqrt(10),9,1-8/sqrt(10))


  )}}}

Now we can sketch in the hyperbola:
 
{{{drawing(400,400,-6,8,-6,8,
 
line(.9,1,1.1,1), line(1,.9,1,1.1), locate(1.2,1.5,"C(1,1)"),
line(1-sqrt(10),1,1,1), locate(-5.7,1.6,V1(1-sqrt(10),1)),
line(1+sqrt(10),1,1,1), locate(4.5,1.6,V2(1+sqrt(10),1)),
line(1,1,1,2), locate(.8,2.7,"(1,2)"), 
line(1,1,1,0), locate(.8,-.8,"(1,0)"),
graph(400,400,-6,8,-6,8), rectangle(1-sqrt(10),0,1+sqrt(10),2),
line(-7,1-8/sqrt(10),9,1+8/sqrt(10)),
line(-7,1+8/sqrt(10),9,1-8/sqrt(10)),
graph(400,400,-6,8,-6,8,1+sqrt((x-1)^2/10-1)), 
graph(400,400,-6,8,-6,8,1-sqrt((x-1)^2/10-1)) 


  )}}}
 
Next we find the equations of the two asymptotes.
Their slopes are ±{{{b/a}}} or ±{{{1/sqrt(10)}}}
 
The asymptote that has slope {{{1/sqrt(10)}}} goes through the center
C(1,1), so its equation is found using the point-slope
formula:
 
{{{y-y[1]=m(x-x[1])}}}
{{{y-(1)=(1/sqrt(10))(x-1))}}}
{{{y-1=(1/sqrt(10))(x-1))}}}

Multiply through by {{{sqrt(10)}}}
{{{sqrt(10)y-sqrt(10)=x-1}}}
{{{-x+sqrt(10)y=sqrt(10)-1}}}
{{{x-sqrt(10)y=-sqrt(10)+1}}}

The asymptote that has slope {{{-1/sqrt(10)}}} goes through the center
C(1,1), so its equation is also found using the point-slope
formula:
 
{{{y-y[1]=m(x-x[1])}}}
{{{y-(1)=(-1/sqrt(10))(x-1))}}}
{{{y-1=(-1/sqrt(10))(x-1))}}}

Multiply through by {{{sqrt(10)}}}
{{{sqrt(10)y-sqrt(10)=-1(x-1)}}}
{{{sqrt(10)y-sqrt(10)=-x+1}}}

{{{x+sqrt(10)y=sqrt(10)+1}}}
{{{x-sqrt(10)y=-sqrt(10)+1}}}
 
All that's left is to find the two foci.

The distance from the vertex, through the center
to each foci is c units. We calculate c from

{{{c^2=a^2+b^2}}}

{{{c^2=(sqrt(10))^2+(1)^2}}}

{{{c^2=(10+1)}}}

{{{c^2=11}}}

{{{c=sqrt(11)}}}

So the left focus is {{{sqrt(11)}}} units
left of the center (1,1), so the left
focus is the point ({{{1-sqrt(11)}}},1).  That's
just a little left of the vertex V1. And the right
focus is {{{sqrt(11)}}} units right of the center
(1,1), so the right focus is the point ({{{1-sqrt(11)}}},1).
That's just a little right of the vertex V2. I won't
bother to plot them.

Edwin</pre>