Question

find the center and vertices for this hyperbola: 4x² - 9y² - 24x - 36y - 36 = 0 center = (?, ) vertices = (, ) and (, )

Explanation:

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

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

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Complete the square:
For \(x^{2}-6x=(x - 3)^{2}-9\), and \(y^{2}+4y=(y + 2)^{2}-4\)

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Divide by \(36\) to get the standard form \(\frac{(x - 3)^{2}}{9}-\frac{(y + 2)^{2}}{4}=1\)

Step2: Identify the center

For a hyperbola of the form \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\), the center is \((h,k)\). Here \(h = 3\) and \(k=-2\)

Step3: Identify the vertices

For a hyperbola \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\), the vertices are \((h\pm a,k)\). Since \(a^{2}=9\), then \(a = 3\)
The vertices are \((3+3,-2)=(6,-2)\) and \((3 - 3,-2)=(0,-2)\)

Answer:

Center \(=(3,-2)\)
Vertices \(=(6,-2)\) and \((0,-2)\)