Question

The equation of a hyperbola is given by .

Answer 1

Do yhu want me to give yhu the answers I have?

Answer 2

center we got right? \((6,0)\) clear?

Answer 3

Yes. c:

Answer 4

how about foci?

Answer 5

Uh... they said (6,0,+-13)

Answer 6

ok once again we have to know what it looks like in advance

Answer 7

Uhm... I wrote that down the last time yhu told me. ;o One sec... I have all my notes right here.

Answer 8

this is different, it in not an ellipse

Answer 9

.... It would be the first one right? Cuz the y is the first letter thing?

Answer 10

yes!

Answer 11

general form is \[\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\] in your case
\[\frac{y^2}{25}-\frac{(x-6)^2}{144}=1\] this makes \((h,k)=(6,5)\) and \(a=5,b=12\)
good so far?

Answer 12

oops wrong
\[(h,k)=(6,0)\]is what i meant
that is the center

Answer 13

I think I got that yea.

Answer 14

ok now since \(a=5\) and we know that it looks like, the vertices are 5 units ABOVE AND BELOW the center (not left and right)
5 above and below \((6,0)\) are the points \((6,5)\)a nd \((6,-5)\) those are the vertices

Answer 15

yhu did the 5^2 thing right? or the square root of 25, that's how yhu got 5?

Answer 16

zactly

Answer 17

So... Those points... Are the foci? Right?

Answer 18

no not the foci, the vertices
we didn't get to the foci yet

Answer 19

for the foci we need \(c^2=a^2+b^2\) which in this case is \(c^2=25+144=169\)

Answer 20

this makes \(c=\sqrt{169}=13\)

Answer 21

we have to go 13 units up and down from the center to get to the foci
that is why they are \((6,13)\) and \((6,-13)\)

Answer 22

Oh, so the points were the points the the hyperbola is going to be graphed on?

Answer 23

the foci are not on the hyperbola, the vertices are

Answer 24

it is almost impossible for me to graph a hyperbola here that doesn’t look like a two year old drew it
lets use technology
http://www.wolframalpha.com/input/?i=hyperbola+y^2%2F25-%28x-6%29^2%2F144%3D1

Answer 25

the vertices are the points here (now i have to draw)