Para poder determinar los elementos de la hipérbola, tendremos que llevar la ecuación que tenemos a la siguiente forma:
[tex]\boxed{\boldsymbol{\sf{\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1 }}}[/tex]
Por ello completaremos cuadrados
[tex]\begin{array}{c}\sf{9x^2 - 16y^2 + 36x + 32y - 124 = 0}\\\\\sf{9x^2 +36x- 16y^2 + 32y - 124 = 0}\\\\\sf{(3x)^2 +2(3x)\boldsymbol{\sf{(6)}}- (4y)^2 + 2(4y)\boldsymbol{\sf{(4)}} - 124 = 0}\\\\\sf{(3x)^2 +2(3x)(6)+\boldsymbol{\sf{(6)^2}}-\boldsymbol{\sf{(6)^2}}- \bigg[(4y)^2 - 2(4y)(4)+\boldsymbol{\sf{(4)^2}}-\boldsymbol{\sf{(4)^2}}\bigg] - 124 = 0}\\\\\sf{(3x+6)^2-\boldsymbol{\sf{36}}- \bigg[(4y-4)^2-\boldsymbol{\sf{16}}\bigg] - 124 = 0}\\\\\end{array}[/tex]
[tex]\begin{array}{c}\sf{(3x+6)^2-36-(4y-4)^2+16-124=0}\\\\\sf{(3x+6)^2-(4y-4)^2-144=0}\\\\\sf{\bigg((3x+6)^2-(4y-4)^2\bigg)\dfrac{1}{144}=\bigg(144\bigg)\dfrac{1}{144}}\\\\\sf{\dfrac{(3x+6)^2}{144}-\dfrac{(4y-4)^2}{144}=1}\\\\\sf{\dfrac{(3(x+2))^2}{144}-\dfrac{(4(y-1))^2}{144}=1}\\\\\sf{\dfrac{9(x+2)^2}{144}-\dfrac{16(y-1)^2}{144}=1}\\\\\end{array}[/tex]
[tex]\begin{array}{c}\sf{\dfrac{\dfrac{9(x+2)^2}{9}}{\dfrac{144}{9}}-\dfrac{\dfrac{16(y-1)^2}{16}}{\dfrac{144}{16}}=1}\\\\\sf{\dfrac{\dfrac{9(x+2)^2}{9}}{16}-\dfrac{\dfrac{16(y-1)^2}{16}}{9}=1}\\\\\boxed{\boxed{\boldsymbol{\sf{\dfrac{(x+2)^2}{4^2}-\dfrac{(y-1)^2}{3^2} = 1 }}}}\end{array}[/tex]
Ya teniendo esto podemos determinar sus elementos fácilmente
A) Determina las coordenadas de los vértices
Los vértices tendrán la siguiente forma: V1 = (h-a,k) y V2 =(h+a,k)
[tex]\begin{array}{c}\sf{\dfrac{(x-\overset{\displaystyle\overset{h}{\downarrow}}{(-2)})^2}{\overset{\displaystyle {4}^2}{\overset{\downarrow}{a}}}-\dfrac{ (y-\overset{\displaystyle\overset{k}{\downarrow}}{1})^2}{\overset{\displaystyle {3}^2}{\overset{\downarrow}{b}}} = 1}\\\\\\\begin{array}{ccccc}\sf{V_1=(-2 + 4,1)}&&&\sf{V_2=(-2 - 4,1)}\\\\\boxed{\boldsymbol{\sf{V_1 = (2,1)}}}&&&\boxed{\boldsymbol{\sf{V_2 = (-6,1)}}}\end{array}\end{array}[/tex]
B) Determinar las ecuaciones de sus asintotas
Las asíntotas tendrán la siguiente forma: [tex]\sf{y_1 - k = -\dfrac{b}{a}(x-h)}[/tex] y [tex]\sf{y_2-k = \dfrac{b}{a}(x-h)}[/tex]
[tex]\begin{array}{c}\sf{\dfrac{(x-\overset{\displaystyle\overset{h}{\downarrow}}{(-2)})^2}{\overset{\displaystyle {4}^2}{\overset{\downarrow}{a}}}-\dfrac{ (y-\overset{\displaystyle\overset{k}{\downarrow}}{1})^2}{\overset{\displaystyle {3}^2}{\overset{\downarrow}{b}}} = 1}\\\\\\\begin{array}{ccccc}\sf{y_1-k = -\dfrac{b}{a}(x-h)}&&&\sf{y_2-k = \dfrac{b}{a}(x-h)}\\\\\boxed{\boldsymbol{\sf{y_1 = -\dfrac{3}{4}(x+2)+1}}}&&&\boxed{\boldsymbol{\sf{y_1 = \dfrac{3}{4}(x+2)+1}}}\end{array}\end{array}[/tex]
C) Determina las coordenadas de los focos de la hipérbola
Los focos tendrán la siguiente forma: F1 = (h-c,k) y F2 =(h+c,k) siendo c² = a² + b², entonces
[tex]\begin{array}{c}\sf{\dfrac{(x+2)^2}{\overset{\displaystyle {4}^2}{\overset{\downarrow}{a}}}-\dfrac{ (y-1)^2}{\overset{\displaystyle {3}^2}{\overset{\downarrow}{b}}} = 1}\\\\\\\sf{c^2=a^2+b^2}\\\\\sf{c^2 = 4^2 + 3^2}\\\\\sf{c= \sqrt{25}}\\\\\sf{c = 5}\\\begin{array}{cccc}\\\sf{F_1=(h-c,k)}&&&\sf{F_2=(h+c,k)}\\\\\boxed{\boldsymbol{\sf{F_1=(-7,1)}}}&&&\boxed{\boldsymbol{\sf{F_2=(3,1)}}}\\\end{array}\end{array}[/tex]
Todos los elementos lo puedes ver en la gráfica.
[tex]\boxed{\sf{{R}}\quad\raisebox{10pt}{$\sf{\red{O}}$}\!\!\!\!\raisebox{-10pt}{$\sf{\red{O}}$}\quad\raisebox{15pt}{$\sf{{G}}$}\!\!\!\!\raisebox{-15pt}{$\sf{{G}}$}\quad\raisebox{15pt}{$\sf{\red{H}}$}\!\!\!\!\raisebox{-15pt}{$\sf{\red{H}}$}\quad\raisebox{10pt}{$\sf{{E}}$}\!\!\!\!\raisebox{-10pt}{$\sf{{E}}$}\quad\sf{\red{R}}}\hspace{-64.5pt}\rule{10pt}{.2ex}\:\rule{3pt}{1ex}\rule{3pt}{1.5ex}\rule{3pt}{2ex}\rule{3pt}{1.5ex}\rule{3pt}{1ex}\:\rule{10pt}{.2ex}[/tex]
