Given the equation:
[tex]9x^2-y^2=9[/tex]
we can divide both sides by 9 to get the following:
[tex]\begin{gathered} \frac{1}{9}(9x^2-y^2=9) \\ \\ \Rightarrow x^2-\frac{y^2}{9}=1 \\ \frac{x^2}{1}-\frac{y^2}{9}=1 \end{gathered}[/tex]
given the general equation of the hyperbola:
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]
we have in this case that:
[tex]\begin{gathered} a^2=1 \\ b^2=9 \\ \Rightarrow a=1 \\ b=3 \end{gathered}[/tex]
in this case, we have that the center is located at the origin, then:
[tex](h,k)=(0,0)[/tex]
then, the vertices are:
[tex]\begin{gathered} (h\pm a,0) \\ \Rightarrow(0\pm1,0) \\ \Rightarrow(1,0)^{} \\ \text{and} \\ (-1,0) \end{gathered}[/tex]
to find the focal points we have the following expression:
[tex]\begin{gathered} (h\pm c,k) \\ \text{where:} \\ c=\sqrt[]{a^2+b^2} \end{gathered}[/tex]
then, we have:
[tex]\begin{gathered} c=\sqrt[]{1+9}=\sqrt[]{10} \\ \Rightarrow(0\pm\sqrt[]{10},0) \\ \Rightarrow(\sqrt[]{10},0) \\ \text{and} \\ (-\sqrt[]{10},0) \end{gathered}[/tex]
to find the fundamental rectangle, we can find the corners with the following expressions:
[tex]\begin{gathered} (0,b),(0,-b) \\ (a,0),,(-a,0) \\ \Rightarrow \\ (0,3),(0-3) \\ (1,0),(-1,0) \end{gathered}[/tex]
finally, we can find the asypmtotes with the folllowing equation:
[tex]\begin{gathered} y=\pm\frac{b}{a}(x-h)+k \\ \Rightarrow y=\pm\frac{3}{1}(x-0)+0 \\ y=\pm3x \end{gathered}[/tex]
therefore, the graph of the hyperbola is:
