Question 669841
given: {{{4x^2-49y^2=196}}}


First get it in standard form, which is either

{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}} if the hyperbola opens right and left, 

or

{{{(y-k)^2/a^2-(x-h)^2/b^2=1}}} if the hyperbola opens upward and downward.

{{{4x^2/196-49y^2/196=1}}}

{{{x^2/49-y^2/4=1}}}

{{{x^2/7^2-y^2/2^2=1}}}

note that:

the denominator of {{{x^2}}} term is greater than that of the {{{y}}} term, so the major axis is horizontal


 Since {{{x^2=(x-0)^2}}} and {{{y^2 = (y-0)^2}}}, then 

{{{h=0}}} and {{{k=0}}}

so, the center is at ({{{h}}}, {{{k}}}) = ({{{0}}}, {{{0}}}). 

 {{{a=7 }}}and {{{b=2}}}, 

 the equation {{{a^2 + b^2 = c^2}}} gives you {{{c^2 = 49 + 4 = 53}}}, 

so {{{c = sqrt(53)= (7.28)}}}


and the eccentricity is {{{e = 7.28/7=1.04}}}

the vertices and foci are above and below the center, so the foci are at

 ({{{0}}},{{{-c}}}) and ({{{0}}},{{{c}}}), which is
 
({{{0}}},{{{-7.28}}}) and ({{{0}}}, {{{7.28}}})
 
and the vertices are at ({{{0}}},{{{a}}}) and ({{{0}}},{{{-a}}}) which is 

({{{0}}},{{{7}}}) and ({{{0}}},{{{-7}}})


Because the {{{x}}} part of the equation is dominant , then the slope of the 

asymptotes has the a on top, so the slopes will be {{{m = (2x/7)=(0.29)}}}. 

so, the asymptotes are {{{y=(0.29)x}}} and {{{y=-(0.29)x}}}


graph:

{{{graph( 600, 600, -15, 15, -10, 10, 4x^2-49y^2<=196 ))}}}

{{{ graph( 600, 600, -15, 15, -10, 10, sqrt(x^2/7-4),-sqrt(x^2/7-4)) }}}