Hi,

x^2-y^2+2x+6y-9=0 

(x+1)^2 -1 - (y-3)^2 +9 - 9 = 0    || In completing Square need to subtract 1 and add 9

  C(-1,3) with V(-2,3) and V(1,3)

   Foci are ( -1 ±  ,3)

Standard Form of an Equation of an Hyperbola opening right and  left is: 

with C(h,k) and vertices 'a' units right and left of center,   2a the length of the transverse axis

Foci are  units right and left of center along y = k

& Asymptotes Lines passing thru C(h,k), with slopes  m =  ± b/a 

 

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
 
Apparently you didn't complete the square correctly:







Half of 2 is 1.  Squared is 1

Add +1 inside the first parentheses and to the right side.

Half of -6 is -3.  Squared is 9.

Add +9 inside the second parentheses and since the second
parentheses is preceded by a - sign we add -9 to the right
side:





That's equivalent to



I think you can go from there.  If not post again.

Edwin
  



  

Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!
 





Keeping the signs straight when completing the square on the general quadratic for a hyperbola can be a little tricky.  Watch carefully:











Move the constant term to the RHS:











Rearrange:











Since you are working a hyperbola, group the "negative variable" terms with parentheses:











Note the sign change on the first degree  term because of the distributive law.





Now complete the square on both variables:











Notice that because you are adding a +9 INSIDE of the parentheses on the left, you are actually adding a -9 to the LHS.  So you need a -9 in the RHS to make up for it.





.





so center at  just like you had it, but , , and therefore 





I think you can take it from here, can you not?





John



My calculator said it, I believe it, that settles it



 

Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
 What are the correct vertices, foci, and asymptotes for the following hyperbola?: 

x^2-y^2+2x+6y-9=0

complete the squares

x^2+2x-y^2+6y-9=0

(x^2+2x+1)-(y^2-6y+9)=9+1-9

(x+1)^2-(y-3)^2=1 (This is where you made a fatal error)

This is an equation of a hyperbola with horizontal transverse axis.

Its standard form:, (h,k)=(x,y) coordinates of center

For given equation:

center: (-1,3)

a^2=1

a=1

vertices: (-1±a,-3)=(-1±1,-3)=(-2,-3) and (0,-3)

..

b^2=1

b=1

..

c^2=a^2+b^2=1+1=2

c=√2≈1.4

Foci: (-1±c,-3)=(-1±1.4,-3)=(-2.4,-3) and (0.4,-3)

..

Asymptotes are straight lines that go thru the center(-1,3)

Standard form of equation for straight lines: y=mx+b, m=slope, b=y-intercept

slopes of asymptotes for hyperbolas: ±b/a=±1/1=±1

Equations for asymptotes:

y=x+b

solve for b using coordinates of the center

3=-1+b

b=4

equation: y=x+4

..

y=-x+b

solve for b using coordinates of the center

3=1+b

b=2

equation: y=-x+2





 

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