SOLUTION: Can ANYONE help me with this problem? No one seems to know the foci of this equation. Please Help!!! (x^2/16)-(y^2/4)= 1 Thank You, Sarah

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Can ANYONE help me with this problem? No one seems to know the foci of this equation. Please Help!!! (x^2/16)-(y^2/4)= 1 Thank You, Sarah      Log On


   

Question 202428: Can ANYONE help me with this problem? No one seems to know the foci of this equation. Please Help!!!

(x^2/16)-(y^2/4)= 1


Thank You,
Sarah
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
This equation is a hyperbola in the form %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1 (this hyperbola opens left/right).


The foci for any hyperbola in the form %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1 are (h+c,k) and (h-c,k) where a%5E2%2Bb%5E2=c%5E2


In this case, h=0, k=0, a=4 (since 4%5E2=16) and b=2 (since 2%5E2=4)


So 4%5E2%2B2%5E2=c%5E2 ---> c%5E2=20 ---> c=sqrt%2820%29=2%2Asqrt%285%29 which means that c=2%2Asqrt%285%29


Now plug in the given h, k, and c values into (h+c,k) and (h-c,k) to get:

and


and simplify to get and


So the foci are and


Note: the foci approximate to (4.4722, 0) and (-4.4722, 0)