SOLUTION: Given: x^2 - y^2 = 8x -2y -13. Find the center, the vertices, the foci, and the asymptotes. Then draw the graph neatly, please.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Given: x^2 - y^2 = 8x -2y -13. Find the center, the vertices, the foci, and the asymptotes. Then draw the graph neatly, please.      Log On


   



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.
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Given: x%5E2+-+y%5E2+=+8x+-+2y+-+13. Find the center, the vertices, the foci, and the asymptotes. Then draw the graph neatly, please
This is a hyperbola because the x%5E2 and the y%5E2 term have opposite
signs.

x%5E2+-+y%5E2+=+8x+-+2y+-+13

Get it like this:

x%5E2+-+8x+-+y%5E2+%2B+2y+=+-13

Put parentheses around the first two terms:

%28x%5E2+-+8x%29+-+y%5E2+%2B+2y+=+-13

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

%28x%5E2+-+8x%29+-+%28y%5E2+-+2y%29+=+-13

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

%28x%5E2+-+8x%2Bred%2816%29%29+-+%28y%5E2+-+2y%29+=+-13%2Bred%2816%29

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:



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

%28x-4%29%5E2-%28y-1%29%5E2=2

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

%28x-4%29%5E2%2F2-%28y-1%29%5E2%2F2=2%2F2

%28x-4%29%5E2%2F2-%28y-1%29%5E2%2F2=1

Compare that to

%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=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



Comparing further:

a%5E2=2 so a=sqrt%282%29

Length of semi-transverse axis = a = sqrt%282%29

So we draw the transverse axis sqrt%282%29 about 1.4 units
to the right and to the left of the center:



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%282%29,1).  

We add "a" to "h" to get the x-coordinate of the right vertex, so 
the right vertex is V(4+sqrt%282%29,1).  They have the same 
y-coordinate as the center. 

b%5E2=2 so b=sqrt%282%29

Length of semi-conjugate axis = b = sqrt%282%29

So we draw the conjugate axis sqrt%282%29 about 1.4 units
above and below the center:



Next we draw the defining rectangle around the two axes:


 
We draw the asymptotes by drawing the extended diagonals of the
defining rectangle:



These asymptotes have slopes %22%22+%2B-+b%2Fa=%22%22+%2B-+sqrt%282%29%2Fsqrt%282%29=%22%22+%2B-+1

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

y-y%5B1%5D=m%28x-x%5B1%5D%29

Finding the equation of the asymptote with the positive slope

y-1=1%28x-4%29
y-1=x-4
y=x-3

Finding the equation of the asymptote with the begative slope

y-1=-1%28x-4%29
y-1=-x%2B4
y=x%2B5

Finally we sketch in the hyperbola:



Edwin