Ecuación de la hipérbola horizontal

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En este apartado se tratarán las hipérbolas horizontales con el centro en un punto genérico $C (x_{0}, y_{0})$ .

El eje focal es paralelo al eje de abscisas, y por lo tanto los focos están en las coordenadas $F^{'} (x_{0} - c, y_{0})$ y $F (x_{0} + c, y_{0})$ .

Aplicando estos focos en la definición general de la hipérbola $\overset{―}{P F} - \overset{―}{P F^{'}} = 2 a$ se obtiene la expresión $\sqrt{(x - x_{0} + c)^{2} + (y - y_{0})^{2}} - \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}} = 2 a$

Al sumar la raíz, y elevando al cuadrado: $\begin{array}{rcl} (\sqrt{(x - x_{0} + c)^{2} + (y - y_{0})^{2}})^{2} & = & (2 a + \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}})^{2} \\ (x - x_{0} + c)^{2} + (y - y_{0})^{2} & = & 4 a^{2} + 4 a \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}} + \\ + (x - x_{0} - c)^{2} + (y - y_{0})^{2} \\ (x - x_{0})^{2} + 2 (x - x_{0}) c + c^{2} + (y - y_{0})^{2} & = & 4 a^{2} + 4 a \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}} + \\ + (x - x_{0})^{2} - 2 (x - x_{0}) c + c^{2} + (y - y_{0})^{2} \end{array}$

Simplificando y dividiendo por cuatro: $\begin{array}{rcl} 4 (x - x_{0}) c & = & 4 a^{2} + 4 a \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}} \\ (x - x_{0}) c & = & a^{2} + a \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}} \end{array}$

Al despejar la raíz y elevar nuevamente al cuadrado: $\begin{array}{rcl} (c (x - x_{0}) - a^{2})^{2} & = & (a \sqrt{(x - x_{0} - c)^{2} + (y - y_{0})^{2}})^{2} \\ c^{2} (x - x_{0})^{2} - 2 a^{2} c (x - x_{0}) + a^{4} & = & a^{2} ((x - x_{0})^{2} + (y - y_{0})^{2} \\ c^{2} (x - x_{0})^{2} - 2 a^{2} c (x - x_{0}) + a^{4} & = & a^{2} ((x - x_{0})^{2} - 2 c (x - x_{0}) + x^{2} + (y - y_{0})^{2}) \\ c^{2} (x - x_{0})^{2} - 2 a^{2} c (x - x_{0}) + a^{4} & = & a^{2} (x - x_{0})^{2} - 2 a c (x - x_{0}) + a^{2} c^{2} + a^{2} (y - y_{0})^{2} \\ c^{2} (x - x_{0})^{2} - a^{2} (x - x_{0})^{2} - a^{2} (y - y_{0})^{2} & = & a^{2} c^{2} - a^{4} \\ (c^{2} - a^{2}) (x - x_{0})^{2} - a^{2} (y - y_{0})^{2} & = & a^{2} (c^{2} - a^{2}) \end{array}$

Se divide entonces por $a^{2} (c^{2} - a^{2})$ para obtener un $1$ a la derecha: $\begin{array}{rcl} \frac{(c^{2} - a^{2}) (x - x_{0})^{2}}{a^{2} (c^{2} - a^{2})} - \frac{a^{2} (y - y_{0})^{2}}{a^{2} (c^{2} - a^{2})} & = & 1 \\ \frac{(x - x_{0})^{2}}{a^{2}} - \frac{(y - y_{0})^{2}}{(c^{2} - a^{2})} & = & 1 \end{array}$

Aplicando la definición $c^{2} = a^{2} + b^{2}$ , $b^{2} = c^{2} - a^{2}$ se sustituye y se llega a la ecuación deseada: $\frac{(x - x_{0})^{2}}{a^{2}} - \frac{(y - y_{0})^{2}}{b^{2}} = 1$

Ejemplo

Hallar la ecuación de la hipérbola de centro $C (2, 3)$ , vértice $A (2, 6)$ y de foco $F (2, 7)$ .

La distancia focal $c = 7 - 3 = 4 a = 6 - 3 = 3$ Con $c^{2} = a^{2} + b^{2}$ , so $b = \sqrt{c^{2} - a^{2}} = \sqrt{16 - 9} = \sqrt{7}$ .

Sustituyendo en $\frac{(x - x_{0}^{2})}{a^{2}} - \frac{(y - y_{0})^{2}}{b^{2}} = 1$ e identificando $C (x_{0}, y_{0})$ , resulta $\frac{(x - 2)^{2}}{9} - \frac{(y - 3)^{2}}{7} = 1$