Question 1189325


check the first choice:

{{{9x^2-16y^2 = 144 }}}

{{{9x^2/144-16y^2/144 = 144/144 }}}

{{{x^2/16-y^2/9 = 144/144 }}}

=> {{{h=0}}}, {{{k=0}}}, {{{a=4}}},{{{b=3}}}

=>  center ({{{h}}}, {{{k}}} ) is at origin 

The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints.

The vertices ({{{h+a}}}, {{{k}}} ), ({{{h-a}}},{{{ k}}} ) are the two bending points of the hyperbola with center ({{{h}}}, {{{k}}} ) and semi-axis {{{a}}}, {{{b}}}.

so,

({{{h+a}}}, {{{k}}} )= ({{{0+4}}},{{{ 0}}} ) =({{{4}}},{{{0}}})

({{{h-a}}}, {{{k }}})=({{{0-4}}},{{{0}}})=({{{-4}}},{{{0}}})

so, vertices lie on x-axis, and {{{y=0}}}->and its transverse axis lies between ({{{4}}},{{{0}}}) and ({{{-4}}},{{{0}}})



For right-left hyperbola the asymptotes are:

{{{y}}}= ±{{{(b/a)(x-h)+k}}}

we need only positive

{{{y= (3/4)(x-0)+0}}}

{{{y= 3x/4}}}

{{{ 3x=4y}}}


this means, your answer is option a. {{{9x^2-16y^2 = 144 }}}