Question

Very confused need help!

Answer 1

Answer:

$\frac{\left(y-1\right)^2}{64}-\frac{\left(x-3\right)^2}{36}=1$ is the right option.

The graph is also attached below.

Step-by-step explanation:

$\frac{\left(y-1\right)^2}{64}-\frac{\left(x-3\right)^2}{36}=1$

As

$Hyperbola\:standard\:equation$

$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1\:\mathrm{\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

$\mathrm{with\:center\:}\textbf{\left(h,\:k\right)},\:\mathrm{\:semi-axis\:\textbf{a}\:and\:semi-conjugate-axis\:\textbf{b}.}$ $\left(h,\:k\right)$ $,$ $semi\:-\:axis\:a\:and\:semi\:-\:conjugate\:-\:axis\:b.\:$

$\mathrm{Rewrite}\:\frac{\left(y-1\right)^2}{64}-\frac{\left(x-3\right)^2}{36}=1\:\mathrm{in\:the\:form\:of\:a\:standard\:hyperbola\:equation}$

$\frac{\left(y-1\right)^2}{8^2}-\frac{\left(x-3\right)^2}{6^2}=1$

$\mathrm{Therefore\:Hyperbola\:properties\:are:}$

$\left(h,\:k\right)=\left(3,\:1\right),\:a=8,\:b=6$

Therefore, $\frac{\left(y-1\right)^2}{64}-\frac{\left(x-3\right)^2}{36}=1$ is the right option.

The graph is also attached below.

Keywords: hyperbola, graph

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