Is it possible in the standard equation of a hyperbola with center at origin to have a=c or a > c , keeping in mind that if a=c then $b^2 = c^2 - a^2 =0$ which means $y^2 / b^2$ would be undefined. And if a > c then $b^2 = c^2 - a^2$ would be negative which is impossible for a square. If there are cases of hyperbolas where a=c or a > c then what are their equations? Are they still described by $x^2/a^2 - y^2/b^2 $?
i am refering for c that is $c^2 = a^2 + b^2$, i am looking for cases where a>c because as i was reading the derivation of the hyperbola equation b was defined such that $b^2 = c^2 - a^2$ which requires c > a or else $b^2$ will be negative (not possible) or will be 0 (then $y^2/b^2$ is undefined). However as i was searching the internet i saw that it's possible in the case of degenerate hyperbola to have a=c so i am confused.
Too long for a comment:
All hyperbolas are described by the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\,.$
Quite obviously, this equation has real solutions only when $$ x^2=a^2\frac{b^2+y^2}{b^2}\ge a^2\,. $$
In your notation $c^2=a^2+b^2$ we can plug $y=c$ into this inequality and write it as $$ c^2\ge b^2\,. $$
In other words: there is no real $a$ that is greater than $c\,.$ The case $a=c$ leads to $b=0\,.$ In that case the equation for the hyperbola is undefined. What might help is to consider the limit $b\to 0\,.$ The hyperbola then converges to the set $$ \{x^2\ge a^2,y=0\}\,. $$