Question 1183543
 

{{{y^2/M - x^2/N = 1}}}, {{{M}}}, {{{N> 0}}} if the center is at the {{{origin}}}

if given  the standard form {{{y^2/a^2-x^2/b^2=1}}}, then foci have the form  ({{{0}}},±{{{c}}}), and the transverse axis is the y-axis

=> {{{c=sqrt(29)}}}
 the foci are at: ({{{0}}} , {{{sqrt(29)}}}) and ({{{0}}},{{{-sqrt(29)}}} )

in your case
{{{M=a^2}}}
{{{N=b^2}}}
the length of the conjugate axis is {{{2a=10}}}=>{{{a=5}}}=>{{{a^2=25}}}
 
{{{c^2=a^2+b^2}}}
{{{b^2=c^2-a^2}}}
{{{b^2=29-25}}}
{{{b^2=4}}}
 
then
{{{M=a^2=25}}}
{{{N=b^2=4}}}

and your equation is:

{{{y^2/25 - x^2/4 = 1}}}

{{{ graph( 600, 600, -15, 15, -15, 25,  sqrt( 25(x^2/4 +1)), -sqrt( 25(x^2/4 +1))) }}}