Question

Help with a graphic hyperbola problem?

http://prntscr.com/52pqsp

Answer 1

and your thoughts?

Answer 2

i dont get how to find the vertices and co-vertices

Answer 3

well, if we have a center, then the ab parts define distances from it to get to the verts

how would you type out a general hyper equation?

Answer 4

in hypers, the verts are definable points, and the coverts are not ... does that make sense?

Answer 5

h,k+a?

Answer 6

\[\pm\frac{(x-h)^2}{a^2}\mp\frac{(y-k)^2}{b^2}=1\]

Answer 7

oh

Answer 8

in this case, we have +y and -x parts soo

\[\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1\]
when y=k, x is not definable ... so lets start with the vertexes that are definable, let x=h
\[\frac{(y-k)^2}{b^2}-\frac{(h-h)^2}{a^2}=1\]
\[\frac{(y-k)^2}{b^2}=1\]
\[(y-k)^2=b^2\]
\[y-k=\pm\sqrt{b^2}\]
\[y=k\pm b\]

Answer 9

to find the undefinable vertexes ... we let y=k, and ignore the subtraction:
\[\frac{(k-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1\]
\[0-\frac{(x-h)^2}{a^2}=1\]
\[\frac{(x-h)^2}{a^2}=1\]
and solve for x

Answer 10

i got (0,4) and (0,-7)

Answer 11

what was our center ?

Answer 12

vertexes are not neccessarily x and y intercepts; but they do intercept the .... forget what they call the axis in this case ... transversal and something else