Question 1204699


The center of a hyperbola is ({{{4}}},{{{0}}}), so ({{{h}}}, {{{k}}})=({{{4}}},{{{0}}}) => {{{h=4}}}, {{{k=0}}}

One of the vertices is ({{{2}}},{{{0}}}), the same ordinate as the center, so we have hyperbola with a horizontal transverse axis.

For a hyperbola with a horizontal transverse axis, the relationship between the center and vertex is as follows:

vertex is at ({{{2}}},{{{0}}})=({{{h-a}}},{{{k}}}) 

so, {{{h-a=2}}}

since {{{h=4}}}, we have

{{{4-a=2}}}

{{{4-2=a}}}

{{{a=2}}}

For a hyperbola with a horizontal transverse axis, the slope of the asymptotes is

{{{m}}}=±{{{b/a}}}

if given slope {{{m=3/2}}} and {{{a=2}}}, we have

{{{b/2=3/2 }}}=>{{{b=3}}}


The equation of the hyperbola in standard form with a horizontal transverse axis is

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

substitute all the components

{{{(x-4)^2/2^2-(y-0)^2/3^2=1}}}

{{{(x-4)^2/4-y^2/9=1}}}=>your answer


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(4,0,.12),circle(2,0,.12),
locate(4,0.5,C(4,0)),locate(2,0.5,V(2,0)),
graph( 600, 600, -10, 10, -10, 10, -(3/2)sqrt(x^2-8x+12),(3/2)sqrt(x^2-8x+12))) }}}