Question

given the ellipse $\frac{(x - 2)^2}{36}+\frac{(y - 5)^2}{16}=1$,
find the center point:
list the vertices (separated by a comma):
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Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse is \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}} = 1\), where \((h,k)\) is the center of the ellipse.
For the given ellipse \(\frac{(x - 2)^{2}}{36}+\frac{(y - 5)^{2}}{16}=1\), by comparing with the standard form, we can see that \(h = 2\) and \(k=5\). So the center of the ellipse is \((2,5)\).

Step2: Find the vertices

Since \(a^{2}=36\), then \(a = 6\). The major axis is parallel to the \(x\) - axis (because \(a^{2}>b^{2}\), here \(a^{2}=36\) and \(b^{2}=16\)).
The vertices of the ellipse \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\) with major axis parallel to the \(x\) - axis are \((h\pm a,k)\)
Substitute \(h = 2\), \(k = 5\) and \(a = 6\) into the formula for vertices.
When we add \(a\) to \(h\): \(x=2 + 6=8\), \(y = 5\), so one vertex is \((8,5)\)
When we subtract \(a\) from \(h\): \(x=2-6=-4\), \(y = 5\), so the other vertex is \((-4,5)\)

Answer:

Center point: \((2,5)\)
Vertices: \((-4,5),(8,5)\)