Question

one of the vertices of the hyperbola (x-3)^2/11-(y-5)^2/11=1

Answer 1

$$
\text{distance from the center of the hyperbola}\\
\text{ to any of the foci is}\\
\sqrt{a^2+b^2}
$$
so, get that value and add it to, in this case to the "h" from (h,k) center coordinates

Answer 2

well a=2 and b= sqrt of 2 so how would I plug that in

Answer 3

btw, I'm guessing you meant to say \(11^2 \) for denominator, no "11" :|

Answer 4

no just 11

Answer 5

hmm, ok, well, then just \(\sqrt{11^2+11^2} \), what does that give you?

Answer 6

sqrt 242 or 15.5

Answer 7

but I don't think that's righhht. Its not in my answers. Im looking for one of the vertices of the hyperbola

Answer 8

ok, what's your center? your (h,k)

Answer 9

center is 1,3

Answer 10

1,3?

Answer 11

yes

Answer 12

hehe, do you know what your (h,k) values are from the equation?

Answer 13

$$
\cfrac{(x-\color{red}{3})^2}{11}-\cfrac{(y-\color{red}{5})^2}{11}=1
$$

Answer 14

yes perfect

Answer 15

well, since the rational that is POSITIVE is the one with the "x" in the numerator, that means the hyperbola is moving HORIZONTALLY

Answer 16

so, your focus would be at \(\huge \color{red}{h}\pm 15.5 \)

Answer 17

yes except remember that (y-5)^2/7 not 11

Answer 18

foci for plural rather :)

Answer 19

well here are my options:
a (5-sqrt11,3)
b.(5-sqrt11,-3)
(3-sqrt11,-5)
d.(3+sqrt11,5)

Answer 20

well... you want the vertices.... hmmm ok

Answer 21

shoot.... that's even simpler.... from the center, which you already have, the distance is THE DENOMINATOR UNDER THE POSITIVE FRACTION :)

Answer 22

Im confused could you show me

Answer 23

the value not squared :), so \(\sqrt{11}\)

Answer 24

$$
\cfrac{(x-\color{red}{3})^2}{(\color{blue}{\sqrt{11}})^2}-\cfrac{(y-\color{red}{5})^2}{(\sqrt{11})^2}=1
$$

Answer 25

that's pretty much the hyperbola equation in standard form, so you want the "a" element, or the blue one in this case

Answer 26

yess so would it 3-sqrt 11 or 5-squrt 11

Answer 27

well, of the 2 fractions in the equation, which one is the POSITIVE one? the one with the "x" in it or the one with the "y" in it?

Answer 28

the one with the x

Answer 29

that means the hyperbola is moving HORIZONTALLY or opening over the "x", and the vertices will be there

Answer 30

so for (h,k) or (x,y) you have (3,5) for your center, so \(h\pm 15.5\)

Answer 31

wait... it wasn't 15.5.... ahemm, 3.3 rather

Answer 32

or \(\sqrt{11}\)

Answer 33

yess

Answer 34

so the answerrr would be A

Answer 35

is it?, what's your "h" value again?

Answer 36

from the center form (h,k) or what is your "x" coordinate for the center for the hyperbola?