Question 202428
This equation is a hyperbola in the form {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}} (this hyperbola opens left/right).



The <a href="http://www.mathwords.com/f/foci_hyperbola.htm">foci</a> for any hyperbola in the form {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}} are (h+c,k) and (h-c,k) where {{{a^2+b^2=c^2}}}



In this case, {{{h=0}}}, {{{k=0}}}, {{{a=4}}} (since {{{4^2=16}}}) and {{{b=2}}} (since {{{2^2=4}}})



So {{{4^2+2^2=c^2}}} ---> {{{c^2=20}}} ---> {{{c=sqrt(20)=2*sqrt(5)}}} which means that {{{c=2*sqrt(5)}}} 



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


*[Tex \LARGE \left(0+2\sqrt{5},0\right)] and *[Tex \LARGE \left(0-2\sqrt{5},0\right)]



and simplify to get *[Tex \LARGE \left(2\sqrt{5},0\right)] and *[Tex \LARGE \left(-2\sqrt{5},0\right)]



So the foci are *[Tex \LARGE \left(2\sqrt{5},0\right)] and *[Tex \LARGE \left(-2\sqrt{5},0\right)]



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