Respuesta :

Hyperbola general formula:

[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]

In this case, the equation is

[tex]\frac{\mleft(x-1\mright)^2}{4}-\frac{(y-2)^2}{9}=1[/tex]

then,

h = 1

k = 2

a = 2

b = 3

Vertices

[tex]\begin{gathered} (h\pm a,k) \\ \text{substituting} \\ (1\pm2,2) \\ (3,2)\text{ and (-1,2)} \end{gathered}[/tex]

Foci

[tex]\begin{gathered} (h\pm c,k) \\ c^2=a^2+b^2 \\ c^2=4+9 \\ c=\sqrt[]{13} \\ \text{substituting} \\ (1\pm\sqrt[]{13},2) \\ (1+\sqrt[]{13},2)\text{ and }(1-\sqrt[]{13},2) \end{gathered}[/tex]

Conjugate axis endpoints

[tex]\begin{gathered} (h,k\pm b) \\ \text{substituting} \\ (1,2\pm3) \\ (1,5)\text{ and }(1,-1) \end{gathered}[/tex]

Asymptotes

[tex]\begin{gathered} y=k\pm\frac{b}{a}(x-h) \\ \text{substituting} \\ y=2\pm\frac{3}{2}(x-1) \\ y=2+\frac{3}{2}(x-1) \\ \text{and} \\ y=2-\frac{3}{2}(x-1) \end{gathered}[/tex]

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