Reduced equation of the vertical hyperbola

We will deal with reduced vertical hyperbolas. In this case, it is the axis of coordinates that corresponds with the focal axis.

The geometric place of the foci is $F^{\prime }(0,-c)$ and $F(0,c)$. Applying the general definition we obtain $${\displaystyle \sqrt{x^2+(y+c{)}^2}-\sqrt{x^2+(y-c{)}^2}=2a}$$

We move the remaining root to the other member, and square it: $${\displaystyle \begin{array}{rcl}(\sqrt{x^2+(y+c{)}^2}{)}^2& =& (2a+\sqrt{x^2+(y-c{)}^2}{)}^2\\ x^2+(y+c{)}^2& =& 4a^2+4a\sqrt{x^2+(y-c{)}^2}+x^2+(y-c{)}^2\\ x^2+y^2+2yc+c^2& =& 4a^2+4a\sqrt{x^2+(y-c{)}^2}+x^2+y^2-2yc+c^2\end{array}}$$

On having simplified, and dividing by four: $${\displaystyle \begin{array}{rcl}4yc& =& 4a^2+4a\sqrt{x^2+(y-c{)}^2}\\ yc& =& a^2+a\sqrt{x^2+(y-c{)}^2}\end{array}}$$ On having cleared the root and having squared again:

$$\begin{array}{rcl}(cy-a^2{)}^2& =& (a+\sqrt{x^2+(y-c{)}^2}{)}^2\\ c^2{y}^2-2a^{2}cy+a^4& =& a^2(x^2+(y-c{)}^2)\\ c^2{y}^2-2a^{2}cy+a^4& =& a^2(x^2+y^2-2cy+c^2)\\ c^2{y}^2-a^2{y}^2-a^2{x}^2& =& a^2{c}^2-a^4\\ (c^2-a^2)y^2-a^2{x}^2& =& a^2(c^2-a^2)\end{array}$$

Then divide using $a^2(c^2-a^2)$ to obtain 1 on the right: $${\displaystyle \begin{array}{rcl}\frac{(c^2-a^2)y^2}{a^2(c^2-a^2)}-\frac{a^2{x}^2}{a^2(c^2-a^2)}& =& 1\\ \frac{y^2}{a^2}-\frac{x^2}{(c^2-a^2)}& =& 1\end{array}}$$

On having applied the definition $c^2=a^2+b^2$, then $b^2=c^2-a^2$ is replaced and goes over to the equation desired for the reduced vertical hyperbola: $${\displaystyle \frac{y^2}{a^2}-\frac{x^2}{b^2}=1}$$