Question 959216
{{{x^2/3-16y^2/12=1}}}
{{{x^2/( sqrt(3) )^2-16y^2/12 = 1}}}
Divide top and bottom of the 2nd term by {{{ 16 }}}
{{{x^2/( sqrt(3) )^2-y^2/((12/16)) = 1}}}
{{{x^2/( sqrt(3) )^2-y^2/((3/4)) = 1}}}
{{{x^2/( sqrt(3) )^2-y^2/((sqrt(3/4))^2) = 1}}}
{{{ a = sqrt(3) }}}
{{{ b = sqrt(3/4) }}}
Hope this makes sense