Question 824656
{{{x^2/a^2-y^2/b^2=1}}};
Vertex A gives {{{sqrt(6)*sqrt(6)/a^2-0^2/b^2=1}}}
{{{6/a^2=1}}}
{{{a=0+- sqrt(6)}}}, just say {{{highlight_green(a=sqrt(6))}}}.


Now knowing equation is {{{x^2/6-y^2/b^2=1}}}, do have enough information to find x when y is known as y=2?  IF not, do you know any way to relate a and b?  Will it help?