Question 43057
{{{y^2= 5x^2+25}}}
or {{{y^2-5x^2= 25}}}
or {{{(y^2-5x^2)/25=25/25)}}}
or {{{y^2/25 - x^2/5 = 1}}}
or {{{y^2/5^2 - x^2/(sqrt(5))^2 = 1}}}


As simple as that! 
Don't make the algebra complex. 
Divide both sides by 25 (to make the term on right side = 1).
This is a hyperbola and not an ellipse.
You derived a wrong equation resulting in wrong type of conic (you got ellipse).


Comparing with standard equation of hyperbola
{{{y^2/a^2 - x^2/b^2 = 1}}} 
where '2a' and '2b' are the lengths of transverse and conjugate axes respectively we have
a = 5, b = {{{sqrt(5)}}}


Now, eccentricity is {{{e=sqrt(1 + b^2/a^2)}}}
or {{{e=sqrt(1 + 5/5^2)}}}
or {{{e=sqrt(1 + 1/5)}}}
or {{{e=sqrt(6/5)}}}


The foci are at (0,ae) and (0,-ae) i.e. (0,{{{5sqrt(6/5)}}}) and (0,{{{-5sqrt(6/5)}}}) i.e. (0,{{{sqrt(30)}}}) and (0,{{{-sqrt(30)}}}).


{{{graph(600,600,-20,20,-20,20, sqrt(5x^2+25), -sqrt(5x^2+25))}}}