Question 379778
Given: {{{x^2 - y^2 = 8x - 2y - 13}}}. Find the center, the vertices, the foci, and the asymptotes. Then draw the graph neatly, please
<pre>
This is a hyperbola because the {{{x^2}}} and the {{{y^2}}} term have opposite
signs.

{{{x^2 - y^2 = 8x - 2y - 13}}}

Get it like this:

{{{x^2 - 8x - y^2 + 2y = -13}}}

Put parentheses around the first two terms:

{{{(x^2 - 8x) - y^2 + 2y = -13}}}

Factor -1 out of the last two terms on the left

{{{(x^2 - 8x) - (y^2 - 2y) = -13}}}

Take half of -8, get -4, square it get +16, add +16 inside the 
first parentheses and add +16 to the right side:

{{{(x^2 - 8x+red(16)) - (y^2 - 2y) = -13+red(16)}}}

Take half of -2, get -1, square it get +1, add +1 inside the second
parentheses, but because of the - in front of the second parentheses
on the left we add -1 to the right side:

{{{(x^2 - 8x+red(16)) - (y^2 - 2y+green(1)) = -13+red(16)-green(1)}}}

Factor the two parentheses as perfect squares, Combine the terms on the
right.

{{{(x-4)^2-(y-1)^2=2}}}

Get a 1 on the right side by dividing every term through by 2

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

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

Compare that to

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

and since the term in x is the positive one, we know the hyperbola 
opens right and left. (As we know the x-axis goes right and left, a
way to remember it).

Comparing further, center = (h,k) = (4,1)

We plot the center

{{{drawing(400,400,-2,10,-5,7, graph(400,400,-2,10,-5,7),

circle(4,1,.1) )}}}

Comparing further:

{{{a^2=2}}} so {{{a=sqrt(2)}}}

Length of semi-transverse axis = a = {{{sqrt(2)}}}

So we draw the transverse axis {{{sqrt(2)}}} about 1.4 units
to the right and to the left of the center:

{{{drawing(400,400,-2,10,-5,7, graph(400,400,-2,10,-5,7),
green(line(4-sqrt(2),1,4+sqrt(2),1)),
circle(4,1,.1) )}}}

The ends of the transverse axis are the vertices. We subtract "a" 
from "h" to get the x-coordinate of the left vertex, so the left
vertex is 

(4-{{{sqrt(2)}}},1).  

We add "a" to "h" to get the x-coordinate of the right vertex, so 
the right vertex is V(4+{{{sqrt(2)}}},1).  They have the same 
y-coordinate as the center. 

{{{b^2=2}}} so {{{b=sqrt(2)}}}

Length of semi-conjugate axis = b = {{{sqrt(2)}}}

So we draw the conjugate axis {{{sqrt(2)}}} about 1.4 units
above and below the center:

{{{drawing(400,400,-2,10,-5,7, graph(400,400,-2,10,-5,7),
green(line(4-sqrt(2),1,4+sqrt(2),1), line(4,1+sqrt(2),4,1-sqrt(2))), 
circle(4,1,.1) )}}}

Next we draw the defining rectangle around the two axes:

{{{drawing(400,400,-2,10,-5,7, graph(400,400,-2,10,-5,7),
green(line(4-sqrt(2),1,4+sqrt(2),1), line(4,1+sqrt(2),4,1-sqrt(2))), 
circle(4,1,.1), red(rectangle(4-sqrt(2),1-sqrt(2),4+sqrt(2),1+sqrt(2))) )}}}
 
We draw the asymptotes by drawing the extended diagonals of the
defining rectangle:

{{{drawing(400,400,-2,10,-5,7, graph(400,400,-2,10,-5,7),
green(line(4-sqrt(2),1,4+sqrt(2),1), line(4,1+sqrt(2),4,1-sqrt(2))), 
circle(4,1,.1), red(rectangle(4-sqrt(2),1-sqrt(2),4+sqrt(2),1+sqrt(2))),

line(-5,-8,16,13), line(-5,10,16,-11) )}}}

These asymptotes have slopes {{{"" +- b/a="" +- sqrt(2)/sqrt(2)="" +- 1}}}

The go through the center (4,1) so their equations are gotten using
the point-slope formula:

{{{y-y[1]=m(x-x[1])}}}

Finding the equation of the asymptote with the positive slope

{{{y-1=1(x-4)}}}
{{{y-1=x-4}}}
{{{y=x-3}}}

Finding the equation of the asymptote with the begative slope

{{{y-1=-1(x-4)}}}
{{{y-1=-x+4}}}
{{{y=x+5}}}

Finally we sketch in the hyperbola:

{{{drawing(400,400,-2,10,-5,7, graph(400,400,-2,10,-5,7,1+sqrt((x-4)^2-2)),
green(line(4-sqrt(2),1,4+sqrt(2),1), line(4,1+sqrt(2),4,1-sqrt(2))), 
circle(4,1,.1), red(rectangle(4-sqrt(2),1-sqrt(2),4+sqrt(2),1+sqrt(2))),
graph(400,400,-2,10,-5,7,1-sqrt((x-4)^2-2)),

line(-5,-8,16,13), line(-5,10,16,-11) )}}}

Edwin</pre>