Question 972995
{{{x^2/16+y^2/64=1}}} ....this is an ellipse, not hyperbola

looking at the denominators, I see that {{{a^2 = 16}}} and {{{b^2 = 64}}}, 

so 
semi-minor axis length {{{a = 4}}}
 and 
semi-major axis length {{{b = 8}}}

the equation {{{c^2 =a^2 +b^2}}} tells me that 

{{{c^2 = 16 + 64 = 80}}}, 

so {{{c =sqrt( 80)=sqrt( 4*4*5)=2*2sqrt( 5)}}}

{{{c=4sqrt( 5)}}}


since {{{(x-0)^2 =x^2}}}  and {{{(y-0)^2=y^2 }}}, then the center is at 

({{{h}}}, {{{k}}}) = ({{{0}}}, {{{0}}}) 


the vertices are at ({{{0}}}, {{{-a}}}) and ({{{0}}},{{{ a}}})
=> ({{{0}}},{{{ -4}}}) and ({{{0}}},{{{ 4}}}),  

and the foci at ({{{0}}},{{{sqrt(b^2-a^2)}}})

{{{sqrt(b^2-a^2)}}}
={{{sqrt(64-16)}}}
={{{sqrt(48)}}}
={{{sqrt(4^2*3)}}}
={{{4sqrt(3)}}}

 ({{{0}}}, {{{-4sqrt(3)}}}) and ({{{0}}}, {{{4sqrt(3)}}}) 


asymptotes:
Ellipses do not have asymptotes. Hyperbolas do, but Ellipses do not.


{{{ graph( 600, 600, -10, 10, -10, 10,sqrt((1-x^2/16)64),-sqrt((1-x^2/16)64)) }}} 



if it’s minus

{{{x^2/16-y^2/64=1}}} then you have hyperbola

and {{{a=4}}}, {{{b=8}}}, {{{c=4sqrt(5)}}}

so, 
foci is at:
 ({{{-4sqrt(5)}}}, {{{0}}})  and ({{{4sqrt(5)}}}, {{{0}}}) 
or ({{{-8.9}}}, {{{0}}})  and ({{{8.9}}}, {{{0}}})

vertices: ({{{-4}}},{{{ 0}}})  and ({{{4}}}, {{{0}}})
center :({{{0}}}, {{{0}}})
semimajor axis length: {{{ 4}}}
semiminor axis length: {{{8}}}

eccentricity :{{{c/a}}}=>{{{4sqrt(5)/4= sqrt(5)}}} or {{{2.2}}}
asymptotes :{{{y = 2 x}}}  and  {{{y = -2 x}}}
 

{{{ graph( 600, 600, -10, 10, -10, 10,sqrt((-1+x^2/16)64),2x,-2x,-sqrt((-1+x^2/16)64)) }}}