For what values of $m$ and $c$ does the ellipse have focii at $(\pm c,0)$?

Question

The ellipse equation is:

$$(1+m^2)x^2+y^2 = 1 + 2m^2 + m^4$$

I used Remarkable identities concepts and distributive rule for development but I got stuck. I considered $a = (1+m^2)^2$ and $b=(1+m^2)$.

$$(1+m^2)x^2+y^2 = 1 + 2m^2 + m^4 $$ $$ (1+m^2)x^2+y^2 = (1+m^2)^2 $$ $$ a^2 = b^2 + c^2 $$ $$ (1+m^2)^2 = (1+m^2) + c^2 $$ $$ 1 + 2m^2 + m^4 = 1 + m^2 + c^2 $$ $$ 1 + 2m^2 + m^4 - 1 - m^2 = c^2 $$ $$ m^4 - m^2 = c^2 $$

Any help?

Answer 1

Hint: Note that the ellipse is vertical, $b^2=1+m^2$ and $a^2=(1+m^2)^2$ then $$c^2=a^2-b^2=m^2(1+m^2)$$