Question

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

Explanation:

Step1: Recall the standard form of an ellipse

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

Step2: Find the center

Since \(h = 4\) and \(k = 5\) in the equation \(\frac{(x - 4)^2}{9}+\frac{(y - 5)^2}{4}=1\), the center of the ellipse is \((4,5)\).

Step3: Recall the formula for vertices

For the ellipse \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (\(a>b\)), the vertices are \((h\pm a,k)\). Here \(a^2 = 9\), so \(a = 3\), \(h = 4\), \(k = 5\).

Step4: Calculate the vertices

Substitute the values into the formula for vertices. When \(x=h + a\), \(x=4 + 3=7\) and \(y = 5\); when \(x=h - a\), \(x=4-3 = 1\) and \(y = 5\). So the vertices are \((1,5)\) and \((7,5)\).

Answer:

Center point: \((4,5)\)
Vertices: \((1,5)\) and \((7,5)\)