1. ## help finding foci

x^2+8y^2+6x-16x+1=0

x^2+6x+9 +8(y^2-2y+1)=-1+9+1=9

(x+3)^2/9 + 8(y-1)^2/9

h=-3

i am supposed to get sqrt 14 but i just dont see how. Can somebody please explain to me why it is supposed to be sqrt 14? Thank You.

2. Originally Posted by blackhawk9876
x^2+8y^2+6x-16x+1=0

x^2+6x+9 +8(y^2-2y+1)=-1+9+1=9

(x+3)^2/9 + 8(y-1)^2/9

h=-3

i am supposed to get sqrt 14 but i just dont see how. Can somebody please explain to me why it is supposed to be sqrt 14? Thank You.
$\displaystyle x^2+6x+9+8(y^2-2y+1)=-1+9+8$

$\displaystyle (x+3)^2+8(y-1)^2=16$

$\displaystyle \frac{(x+3)^2}{16}+\frac{(y-1)^2}{2}=1$

$\displaystyle center \ \ = \ \ (h, k) \ \ = \ \ (-3, 1)$

$\displaystyle a^2 = 16$
$\displaystyle a=4$

$\displaystyle b^2 = 2$
$\displaystyle b = \sqrt2$

$\displaystyle c^2=a^2-b^2$

Can you finish?

The Following is copied from Jhevon's referenced post. This should finish things up nicely. Thanks Jhevon!!

<Quote=Jhevon>

The equation of an ellipse with a horizontal major axis can be expressed as

$\displaystyle \frac {(x - h)^2}{a^2} + \frac {(y - k)^2}{b^2} = 1 \mbox { for } \boxed {a \geq b > 0}$

When in this form, the following hold:

Center: $\displaystyle (h,k)$
Vertices: $\displaystyle (h \pm a,k)$
Foci: $\displaystyle (h \pm c, k)$
........where $\displaystyle c^2 = a^2 - b^2$

<End Quote = Jhevon>

3. For reference, see post #2 here

4. Yes I can finish, thank you. I kept adding 1 instead of 8 thank you. Funny how you can get so stumped and its just a simple error messing the whole problem up

5. Originally Posted by blackhawk9876
Yes I can finish, thank you. I kept adding 1 instead of 8 thank you. Funny how you can get so stumped and its just a simple error messing the whole problem up
yeah, hehe, it's funny afterwards ...frustrating during