Question

Find the Foci of the following graph. Attached! Medal will be given!

Answer 1

hello again.
more conic sections i see

Answer 2

it is hard for me to see, but it there a little yellow box in there?

Answer 3

where does it cross the \(y\) axis?

Answer 4

ok got it

Answer 5

this means the equation of the hypebola is
\[\frac{x^2}{2^2}-\frac{y^2}{3^2}=1\]or
\[\frac{x^2}{4}-\frac{y^2}{9}=1\] focus will be
\[\sqrt{4+9}=\sqrt{13}\] units to the left and right of the center, which in your case is the origin
answer is TRUE

Answer 6

yw
aren't you done with these conics yet?

Answer 7

you need help now? or are you done?

Answer 8

well let me try to explain
if it is oriented the way of the first graph the \(x^2\) term comes first
for this one the \(y^2\) term comes first

Answer 9

so it will look like
\[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\]

Answer 10

you get \(a\) and \(b\) from the picture

Answer 11

since it crosses the \(y\) axis at \(3\) you know \(a=3\) and of course \(a^2=9\)

Answer 12

then you look at the yellow box
since it crosses the \(x\) axis at 5, \(b=5\) and \(b^2=25\)
your equation is therefore \(\frac{y^2}{9}-\frac{x^2}{25}=1\)

Answer 13

ok good
now the foci will be \(\sqrt{25+9}=\sqrt{34}\) units above and below the center, which again is the origin
therefore they will be at \((0,-\sqrt{34})\) and \((0,\sqrt{34})\)

Answer 14

lets check it
http://www.wolframalpha.com/input/?i=hyperbola+y^2%2F9-x^2%2F25%3D1

Answer 15

but really you have only a graph to work with, so you do the computation mostly by your eyeballs
you see \(a=3\) so \(a^2=9\) and \(b=5\) so \(b^2=25\) and so on
if you have another i will be happy to help

Answer 16

yw, good luck and post more if you need more checking