Question 1194068


The standard equation for a hyperbola with a horizontal transverse axis is 

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

given:

 center at ({{{0}}},{{{0}}})=>{{{h=0}}} and {{{k=0}}}


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


its straight side is {{{6}}} units : a latus rectum (literally, “straight side”).
 
 the latus rectum of a hyperbola is {{{(2b^2)/a}}}

{{{(2b^2)/a=6}}}


its eccentricity is {{{e=sqrt(7)/2=c/a }}}=>{{{c=sqrt(7)}}} and{{{ a=2}}}

substitute {{{a}}} in latus rectum formula

{{{(2b^2)/2=6}}}

{{{2b^2=12}}}

{{{b^2=6}}}

then your equation is:

{{{x^2/4 -  y^2/6= 1}}}


{{{ graph( 600, 600, -10, 10, -10, 10,-sqrt(3/2)sqrt(x^2 - 4),sqrt(3/2)sqrt(x^2 - 4) ) }}}