help finding foci

**blackhawk9876** · June 25th 2008, 01:04 PM

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.

**masters** · June 25th 2008, 02:29 PM

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.

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

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

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

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

$a^2 = 16$
$a=4$

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

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

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

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

When in this form, the following hold:

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

<End Quote = Jhevon>

**Jhevon** · June 25th 2008, 02:32 PM

For reference, see post #2 here

**blackhawk9876** · June 26th 2008, 08:00 AM

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

**Jhevon** · June 26th 2008, 10:26 AM

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

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