Question

given the ellipse $\frac{(x - 5)^2}{9}+\frac{(y - 4)^2}{36}=1$
find the center point:
list the four vertices/co - vertices:

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse is \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (\(a>b\)), where \((h,k)\) is the center.
For the given ellipse \(\frac{(x - 5)^2}{9}+\frac{(y - 4)^2}{36}=1\), we have \(h = 5\), \(k=4\).

Step2: Find the vertices and co - vertices

Since \(a^2 = 36\), then \(a = 6\), and \(b^2=9\), then \(b = 3\).
The vertices are \((h,k\pm a)\) and the co - vertices are \((h\pm b,k)\).
Substitute \(h = 5\), \(k = 4\), \(a = 6\), \(b = 3\) into the formulas.
The vertices: \((5,4 + 6)=(5,10)\) and \((5,4-6)=(5,-2)\)
The co - vertices: \((5 + 3,4)=(8,4)\) and \((5-3,4)=(2,4)\)

Answer:

Center point: \((5,4)\)
Four vertices/co - vertices: \((5,10)\), \((5,-2)\), \((8,4)\), \((2,4)\)