Question

match the graphs to their equations
\\( \frac{(y - 1)^{2}}{9}-\frac{(x - 8)^{2}}{1}=1 \\)
\\( \frac{(y + 1)^{2}}{9}-\frac{(x - 8)^{2}}{1}+1 \\)
\\( \frac{(x - 8)^{2}}{9}-\frac{(y - 1)^{2}}{1}=1 \\)
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Explanation:

Step1: Analyze the standard form of hyperbola

The standard form of a hyperbola is \(\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1\) (opens up and down) and \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) (opens left and right).

For the equation \(\frac{(y - 1)^2}{9}-\frac{(x - 3)^2}{1}=1\), it is of the form \(\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1\) with center \((h,k)=(3,1)\), \(a = 3\), \(b = 1\), and it opens up and down. Looking at the graphs, graph \(a\) has a hyperbola that opens up and down with a center around \((3,1)\).

For the equation \(\frac{(y + 1)^2}{9}-\frac{(x - 3)^2}{1}=1\), it is of the form \(\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1\) with center \((h,k)=(3,-1)\), \(a = 3\), \(b = 1\), and it opens up and down. Graph \(c\) has a hyperbola that opens up and down with a center around \((3,-1)\).

For the equation \(\frac{(x - 3)^2}{9}-\frac{(y - 1)^2}{1}=1\), it is of the form \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) with center \((h,k)=(3,1)\), \(a = 3\), \(b = 1\), and it opens left and right. Graph \(d\) has a hyperbola that opens left and right with a center around \((3,1)\).

Answer:

\(\frac{(y - 1)^2}{9}-\frac{(x - 3)^2}{1}=1\) matches graph \(a\), \(\frac{(y + 1)^2}{9}-\frac{(x - 3)^2}{1}=1\) matches graph \(c\), \(\frac{(x - 3)^2}{9}-\frac{(y - 1)^2}{1}=1\) matches graph \(d\)