Question 1204680


The standard form of the equation of a hyperbola with center ({{{0}}},{{{0}}}) and transverse axis on the {{{x}}}-axis is

{{{x^2/a^2-y^2/b^2=1}}}

we know that the coordinates of the center are

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

and  the vertices are 

({{{-a}}},{{{0}}})=({{{-2}}},{{{0}}}) and ({{{a}}},{{{0}}})=({{{2}}},{{{0}}})
=> {{{a=2}}}

since the transverse axis is horizontal, asymptote formula will be:
 
{{{y}}}=±{{{(b/a)x}}}

from graph we see that one asymptote passes through the origin ({{{0}}},{{{0}}}) and ({{{2}}},{{{-1}}})

{{{y=mx}}}

slope is:

{{{m=(-1-0)/(2-0)=-1/2}}}

equation is:

{{{y=-(1/2)x}}}

the other asymptote passes through origin and point ({{{2}}},{{{1}}})

slope is:

{{{m=(1-0)/(2-0)=1/2}}}

equation is:

{{{y=(1/2)x}}}

then, since {{{y}}}=±{{{(b/a)x}}}, {{{b=1}}}

and your equation of hyperbola is:

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

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

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


For the hyperbola, determine the

a) coordinates of the center

({{{0}}},{{{0}}})

b) directions and lengths of both axes

the transverse axis is horizontal
 
the length of the axes is {{{2a =2*2=4}}} and the length of conjugate axis is {{{2b=2*1=2}}}


c) coordinates of the vertices

({{{-2}}},{{{0}}}) and ({{{2}}},{{{0}}}) 


d) slopes of the asymptotes


from graph we see that one asymptote passes through the origin ({{{0}}},{{{0}}})  and ({{{2}}},{{{-1}}}) 

{{{y=mx}}}

slope is:

{{{m=(-1-0)/(2-0)=-1/2}}}

equation is:

{{{y=-(1/2)x}}}

the other asymptote passes through origin and point ({{{2}}},{{{1}}}) 

slope is:

{{{m=(1-0)/(2-0)=1/2}}}

equation is:

{{{y=(1/2)x}}}


{{{ graph( 600, 600, -10, 10, -10, 10, -(1/2)sqrt(x^2-4), sqrt(x^2-4)/2,-(1/2)x,(1/2)x) }}}