Question

given the ellipse $\frac{(x - 5)^{2}}{4}+\frac{(y - 3)^{2}}{16}=1$,
find the center point:
list the vertices:
question help: video written example

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) (\(a>b>0\)), where \((h,k)\) is the center of the ellipse.
For the given ellipse \(\frac{(x - 5)^{2}}{4}+\frac{(y - 3)^{2}}{16}=1\), we can directly identify \(h = 5\) and \(k=3\).

Step2: Find the vertices

Since \(a^{2}=16\), then \(a = 4\). The vertices of the ellipse \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) are \((h,k\pm a)\).
Substitute \(h = 5\), \(k = 3\) and \(a = 4\) into the formula for vertices.
The vertices are \((5,3 + 4)\) and \((5,3-4)\).

Answer:

Center point: \((5,3)\)
Vertices: \((5,7)\) and \((5,-1)\)