Question 112893
find the vertices and foci of the hyperbola 

{{{x^2/144-y^2/9=1}}}

If a hyperbola opens in the horizontal direction, the equation of a hyperbola in standard form appears as

{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}

since {{{a^2=144}}} 

 {{{b^2=9}}} 

=> {{{a}}} = +-{{{12}}}

{{{b}}} = +-{{{3}}}

then we have

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

focal length: 

{{{f}}}=+-{{{ (sqrt (144+9 )) }}}

 and  foci: ({{{0}}},{{{12.37}}}) ({{{0}}},{{{-12.37}}})

the vertices:

since {{{a^2=144}}}=> {{{a}}}= +-{{{12}}}, then the vertices will be at:

({{{-12}}},{{{0}}}) ({{{12}}},{{{0}}})


2.write an equation for the hyperbola. its center is (0,0) 

foci:({{{0}}},{{{-9}}}),({{{0}}},{{{9}}}) 

vertices:({{{0}}},{{{-7}}}),({{{0}}},{{{7}}})
 
since center ({{{h}}},{{{k}}}) is at  ({{{0}}},{{{0}}}), 

=> this is a hyperbola opens in the horizontal direction, the equation of a hyperbola in standard form will be

{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}


since foci are at ({{{0}}},{{{-9}}}), ({{{0}}},{{{9}}}) 

=> {{{a}}} = +-{{{9}}}

=>{{{a^2=81}}} and  

vertices are at ({{{0}}},{{{-7}}}),({{{0}}},{{{7}}}) => 
{{{b= +-7}}} =>{{{b^2=49}}}

then your equation will be:

{{{x^2/81-y^2/49=1}}}