Question

graph the hyperbola given by ( 9 x ^ { 2 } - y ^ { 2 } - 72 x - 4 y + 131 = 0 ).

Explanation:

Step1: Complete the square for \(x\) and \(y\) terms

Group \(x\) - terms and \(y\) - terms:

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For the \(x\) - part: \(x^{2}-8x=(x - 4)^{2}-16\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a = x\), \(b = 4\)). For the \(y\) - part: \(y^{2}+4y=(y + 2)^{2}-4\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=y\), \(b = 2\))

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Divide by \(9\) to get the standard form \(\frac{(x - 4)^{2}}{1}-\frac{(y + 2)^{2}}{9}=1\)

Step2: Identify the center, \(a\), \(b\)

The standard form of a hyperbola is \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\), where \((h,k)\) is the center. Here \(h = 4\), \(k=-2\), \(a = 1\), \(b = 3\)
The center of the hyperbola is \((4,-2)\), and it opens left - right (since the \(x\) - term is positive)

Answer:

The first graph (the one with the center near \((4,-2)\) and opening left - right) is the correct graph of the hyperbola \(9x^{2}-y^{2}-72x - 4y+131 = 0\)