Question 928231
find equation of hyperbola where foci is ({{{9}}},{{{0}}}) and hyperbola passes through ({{{12}}},{{{3}}})



The general formula of a hyperbola is:

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


Since, ({{{12}}}, {{{3}}}) is a point on the hyperbola, we have:


{{{12^2 / a^2 - 3^2 / b^2 = 1}}}

{{{144 / a^2 - 9 / b^2 = 1}}}........eq.(1)


Since the foci is ({{{9}}},{{{0}}}), we have:


{{{a^2 + b^2 = 9^2}}}


{{{b^2 = 81 - a^2}}}.....eq.(2)


Substituting (2) into (1), we have:


{{{144 / a^2 - 9 / (81-a^2) = 1}}} ....common denominator is {{{a^2(81-a^2)}}}


{{{(144(81-a^2)-9a^2)/(a^2(81-a^2))=1}}}


{{{(144(81-a^2)-9a^2)=1(a^2(81-a^2))}}}


{{{11664-144a^2-9a^2=81a^2-a^4}}}


{{{0=144a^2+81a^2+9a^2-a^4-11664}}}


{{{0=234a^2-a^4-11664}}}


{{{a^4-234a^2+11664=0}}}....write {{{-234a^2}}} as {{{-162a^2-72x^2}}}


{{{a^4-162a^2-72x^2+11664=0}}}


{{{(a^4-162a^2)-(72x^2+11664)=0}}}


{{{a^2(a^2-162)-72(x^2+162)=0}}}


{{{(a^2-162)(a^2-72) = 0}}}

solutions:

=>{{{(a^2-162) = 0}}}=>{{{a^2 = 162}}}

=>{{{(a^2-72) = 0}}}=>{{{a^2 = 72}}}


Suppose {{{a^2 = 162}}}, from (2), we have:


{{{b^2 = 81 - 162 = -81}}}=>rejected, since a square cannot be negative 


Suppose {{{a^2 = 72}}}, from (2), we have:

{{{b^2 = 81-72 = 9}}}=> this solution is accepted


Thus, we have:


{{{x^2/72 -y^2/9 = 1}}}

foci is ({{{9}}},{{{0}}}) and ({{{-9}}},{{{0}}})

hyperbola passes through ({{{12}}},{{{3}}})


{{{drawing( 600, 600, -25, 25, -25,25,locate(12,3,p(12,3)), circle(12,3,.3),circle(-9,0,.3),circle(9,0,.3),graph( 600, 600, -25, 25, -25, 25,sqrt(9x^2/72-9),-sqrt(9x^2/72-9))) }}}