A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
find the standard form of the hyperbola with v( plus minus 1, 0) through point (2, 13)
anonymous
 5 years ago
find the standard form of the hyperbola with v( plus minus 1, 0) through point (2, 13)

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0vertices are (1,0), and (1,0); its a convenient thing this is along the x axis i spose

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(2)^2  (13)^2 = 1 is our goal i think

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04 196    = 1 hmmm a^2 b^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0a^2 = 1 since 1/1 = 1 for out vertice

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04  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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol.... 13^2 = 169 ..... i always gets those confused

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^2  \frac{3y^2}{169}=1\] should be better
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.