Question

find the center and vertices for this hyperbola: 9y^2 - 25x^2 - 100x - 108y - 1 = 0 center = (?, ) vertices = (, ) and (, )

Explanation:

Step1: Rearrange and complete the square for x and y terms

Group the x - terms and y - terms:
\[9(y^{2}-12y)-25(x^{2} + 4x)-1=0\]
Complete the square for \(y^{2}-12y=(y - 6)^{2}-36\) and \(x^{2}+4x=(x + 2)^{2}-4\).
\[9((y - 6)^{2}-36)-25((x + 2)^{2}-4)-1=0\]
\[9(y - 6)^{2}-324-25(x + 2)^{2}+100 - 1=0\]
\[9(y - 6)^{2}-25(x + 2)^{2}=225\]
\[\frac{(y - 6)^{2}}{25}-\frac{(x + 2)^{2}}{9}=1\]

Step2: Identify the center

For a hyperbola of the form \(\frac{(y - k)^{2}}{a^{2}}-\frac{(x - h)^{2}}{b^{2}}=1\), the center is \((h,k)\). Here \(h=-2\) and \(k = 6\), so the center is \((-2,6)\).

Step3: Identify the vertices

For a hyperbola of the form \(\frac{(y - k)^{2}}{a^{2}}-\frac{(x - h)^{2}}{b^{2}}=1\), the vertices are \((h,k\pm a)\). Since \(a^{2}=25\), then \(a = 5\), \(h=-2\) and \(k = 6\).
The vertices are \((-2,6 + 5)=(-2,11)\) and \((-2,6-5)=(-2,1)\)

Answer:

Center = \((-2,6)\)
Vertices = \((-2,11)\) and \((-2,1)\)