Question

1. use the standard hyperbola equation to find the y-values to the nearest whole number from the x-values in the table.

\\( \frac { ( x - 2 ) ^ { 2 } } { 4 ^ { 2 } } - \frac { ( y - 1 ) ^ { 2 } } { 2 ^ { 2 } } = 1 \\)
\\( \begin{array} { r r } { x } & { y } \\ { - 7 } & { - } \\ { - 7 } & { - } \\ { 1 } & { } \\ { 5 } & { } \\ { - 5 } & { } \\ { - 3 } & { } \\ { 3 } & { } end{array} \\)

Explanation:

Step1: Substitute \(x = - 7\) into the hyperbola equation

Given \(\frac{(x - 2)^{2}}{4^{2}}-\frac{(y - 1)^{2}}{2^{2}}=1\), when \(x=-7\), we have \(\frac{(-7 - 2)^{2}}{16}-\frac{(y - 1)^{2}}{4}=1\).
First, calculate \(\frac{(-9)^{2}}{16}=\frac{81}{16}\). Then the equation becomes \(\frac{81}{16}-\frac{(y - 1)^{2}}{4}=1\).

Step2: Solve for \((y - 1)^{2}\)

Rearrange the equation: \(\frac{(y - 1)^{2}}{4}=\frac{81}{16}-1=\frac{81 - 16}{16}=\frac{65}{16}\).
Multiply both sides by \(4\): \((y - 1)^{2}=\frac{65}{4}=16.25\).

Step3: Solve for \(y\)

Take the square - root: \(y-1=\pm\sqrt{16.25}\approx\pm4.03\).
Then \(y = 1\pm4.03\). So \(y_1=1 + 4.03\approx5\) and \(y_2=1-4.03\approx - 3\).

Answer:

When \(x=-7\), the \(y\) - values are \(5\) and \(-3\).