Equation of the horizontal hyperbolas

In this part we will deal with horizontal hyperbolas with the center in a generic point $$C(x_0,y_0)$$.

The focal axis is parallel to the abscissa axis, and therefore the foci are in the coordinates $$F'(x_0-c,y_0)$$ and $$F(x_0+c,y_0)$$.

Applying these foci to the general definition of the hyperbola $$\overline{PF}- \overline {PF'}=2a$$ we obtain the expression $$$\sqrt{(x-x_0+c)^2+(y-y_0)^2}-\sqrt{(x-x_0-c)^2+(y-y_0)^2}=2a$$$

On having added the root, and raising to the square: $$$\begin{array}{rcl} \Big(\sqrt{(x-x_0+c)^2+(y-y_0)^2}\Big)^2 & = & \Big(2a+\sqrt{(x-x_0-c)^2+(y-y_0)^2}\Big)^2 \\ (x-x_0+c)^2+(y-y_0)^2 & = & 4a^2+4a\sqrt{(x-x_0-c)^2+(y-y_0)^2}+ \\ & & +(x-x_0-c)^2+(y-y_0)^2 \\ (x-x_0)^2+2(x-x_0)c+c^2+(y-y_0)^2 & = & 4a^2+4a\sqrt{(x-x_0-c)^2+(y-y_0)^2}+ \\ & & +(x-x_0)^2-2(x-x_0)c+c^2+(y-y_0)^2\end{array}$$$

Simplifying and dividing by four: $$$\begin{array}{rcl} 4(x-x_0)c & = & 4a^2+4a \sqrt{(x-x_0-c)^2+(y-y_0)^2} \\ (x-x_0)c & = & a^2+a\sqrt{(x-x_0-c)^2+(y-y_0)^2}\end{array}$$$

On having cleared the root and having squared again: $$$\begin{array}{rcl} (c(x-x_0)-a^2)^2 & = & (a\sqrt{(x-x_0-c)^2+(y-y_0)^2})^2 \\ c^2(x-x_0)^2-2a^2c(x-x_0)+a^4& = & a^2((x-x_0)^2+(y-y_0)^2 \\ c^2(x-x_0)^2-2a^2c(x-x_0)+a^4& = & a^2((x-x_0)^2-2c(x-x_0)+x^2+(y-y_0)^2) \\ c^2(x-x_0)^2-2a^2c(x-x_0)+a^4 & = & a^2(x-x_0)^2-2ac(x-x_0)+a^2c^2+a^2(y-y_0)^2\\ c^2(x-x_0)^2 -a^2(x-x_0)^2-a^2(y-y_0)^2 & = & a^2c^2-a^4 \\ (c^2-a ^2)(x-x_0)^2-a^2(y-y_0)^2& = &a^2(c^2-a^2) \end{array}$$$

We divide by $$a^2(c^2-a^2)$$ to obtain $$1$$ on the right: $$$\begin{array}{rcl} \dfrac{(c^2-a ^2)(x-x_0)^2}{a^2(c^2-a^2)}-\dfrac{a^2(y-y_0)^2}{a^2(c^2-a^2)}& = &1 \\ \dfrac{(x-x_0)^2}{a^2}-\dfrac{(y-y_0)^2}{(c^2-a^2)}& = &1 \end{array}$$$

Applying the definition $$c^2=a^2+b^2$$, $$b^2=c^2-a^2$$ is replaced and we get the desired equation: $$$\displaystyle \frac{(x-x_0)^2}{a^2}-\frac{(y-y_0)^2}{b^2}= 1$$$