Question

find the center and vertices for this hyperbola: 4y² - 49x² - 392x + 16y - 964 = 0 center = (?, ) vertices = (, ) and (, )

Explanation:

Step1: Group the terms and complete the square

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

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Complete the square for \(y\): \(y^{2}+4y=(y + 2)^{2}-4\). Complete the square for \(x\): \(x^{2}+8x=(x + 4)^{2}-16\)

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Divide by \(196\) to get the standard form \(\frac{(y + 2)^{2}}{49}-\frac{(x + 4)^{2}}{4}=1\)

Step2: Identify the center and vertices

For a hyperbola of the form \(\frac{(y - k)^{2}}{a^{2}}-\frac{(x - h)^{2}}{b^{2}}=1\), the center is \((h,k)\), \(a = 7\), \(b = 2\)
The center \((h,k)=(-4,-2)\)
The vertices are \((h,k\pm a)\), so \(k + a=-2+7 = 5\) and \(k - a=-2-7=-9\)

Answer:

Center \(=(-4,-2)\)
Vertices \(=(-4,5)\) and \((-4,-9)\)