Question 1183543
<pre> I think she confused the conjugate axis and the transverse axis. It's
easy to do.

{{{y^2/a^2-x^2/b^2=1}}} is a hyperbola that opens right and left.
<pre>b is the length of the semi-conjugate axis, so b = half of 10, so b = 5. 

{{{y^2/a^2-x^2/5^2=1}}}
{{{y^2/a^2-x^2/25=1}}}

a is the semi-transverse axis, which is the distance from the center to the vertex,</pre>the foci are sqrt(29) units from the center.<pre>
c = the distance from the center to either of the foci.

The Pythagorean relation for all hyperbolas is {{{c^2=a^2+b^2}}}

{{{(sqrt(29))^2=a^2+5^2}}}

{{{29=a^2+25}}}

{{{4=a^2}}}

{{{2=a}}}

The equation is

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

{{{drawing(400,400,-10,10,-10,10,graph(400,400,-10,10,-10,10,(2sqrt(x^2 + 25))/5),graph(400,400,-10,10,-10,10,(-2sqrt(x^2 + 25))/5) )}}} 

Edwin</pre>