Answer:
[tex]\frac{\left(y-1\right)^2}{64}-\frac{\left(x-3\right)^2}{36}=1[/tex] is the right option.
The graph is also attached below.
Step-by-step explanation:
[tex]\frac{\left(y-1\right)^2}{64}-\frac{\left(x-3\right)^2}{36}=1[/tex]
As
[tex]Hyperbola\:standard\:equation[/tex]
[tex]\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1\:\mathrm{\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}[/tex]
[tex]\mathrm{with\:center\:}\textbf{\left(h,\:k\right)},\:\mathrm{\:semi-axis\:\textbf{a}\:and\:semi-conjugate-axis\:\textbf{b}.}[/tex] [tex]\left(h,\:k\right)[/tex] [tex],[/tex] [tex]semi\:-\:axis\:a\:and\:semi\:-\:conjugate\:-\:axis\:b.\:[/tex]
[tex]\mathrm{Rewrite}\:\frac{\left(y-1\right)^2}{64}-\frac{\left(x-3\right)^2}{36}=1\:\mathrm{in\:the\:form\:of\:a\:standard\:hyperbola\:equation}[/tex]
[tex]\frac{\left(y-1\right)^2}{8^2}-\frac{\left(x-3\right)^2}{6^2}=1[/tex]
[tex]\mathrm{Therefore\:Hyperbola\:properties\:are:}[/tex]
[tex]\left(h,\:k\right)=\left(3,\:1\right),\:a=8,\:b=6[/tex]
Therefore, [tex]\frac{\left(y-1\right)^2}{64}-\frac{\left(x-3\right)^2}{36}=1[/tex] is the right option.
The graph is also attached below.
Keywords: hyperbola, graph
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