Question

find the standard form of the hyperbola with v( plus minus 1, 0) through point (2, 13)

Answer 1

vertices are (-1,0), and (1,0); its a convenient thing this is along the x axis i spose

Answer 2

(2)^2 - (13)^2 = 1 is our goal i think

Answer 3

4 196
--- - --- = 1 hmmm
a^2 b^2

Answer 4

a^2 = 1 since 1/1 = 1 for out vertice

Answer 5

4 - 196/b^2 = 1

4b^2 -196 = b^2

3b^2 -196 = 0

b^2 = 196/3

x^2 - 3y^2/196 = 1 is te best i can do

Answer 6

lol.... 13^2 = 169 ..... i always gets those confused

Answer 7

\[x^2 - \frac{3y^2}{169}=1\]
should be better