## animalsavior94 Group Title find the standard form of the hyperbola with v( plus minus 1, 0) through point (2, 13) 3 years ago 3 years ago

1. amistre64 Group Title

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

2. amistre64 Group Title

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

3. amistre64 Group Title

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

4. amistre64 Group Title

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

5. amistre64 Group Title

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

6. amistre64 Group Title

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

7. amistre64 Group Title

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