SOLUTION: What are the correct vertices, foci, and asymptotes for the following hyperbola?: x^2-y^2+2x+6y-9=0 I completed the squares and ended up with this: (x+1)^2/(19)

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Question 629704: What are the correct vertices, foci, and asymptotes for the following hyperbola?:
x^2-y^2+2x+6y-9=0
I completed the squares and ended up with this:
(x+1)^2/(19) - (y-3)^2/(19)=1
But when I plug in (h-a,k)(h+a,k) the answer is not correct for the vertices:
(-1-√19)(-1+√19)
or
(-1-√38,3)(-1+√38,3)for the foci.
What am I doing wrong?

Found 4 solutions by ewatrrr, Edwin McCravy, solver91311, lwsshak3:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!

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(20059)   (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

SOLUTION: What are the correct vertices, foci, and asymptotes for the following hyperbola?: x^2-y^2+2x+6y-9=0 I completed the squares and ended up with this: (x+1)^2/(19) - (y-3)^2/(1