Question

assignment 10.1: the ellipse
score: 20/100 answered: 1/5
question 2
given the ellipse \\( \frac{(x - 5)^2}{16}+\frac{(y - 3)^2}{4}=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 \(\frac{(x - 5)^2}{16}+\frac{(y - 3)^2}{4}=1\), by comparing with the standard form, we can see that \(h = 5\) and \(k=3\). So the center of the ellipse is \((5,3)\).

Step2: Find the vertices

Since \(a^2=16\), then \(a = 4\). The major axis is parallel to the \(x\) - axis (because \(a^2>b^2\) where \(b^2 = 4\)).
The vertices of an ellipse \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) are \((h\pm a,k)\).
Substitute \(h = 5\), \(a = 4\), and \(k = 3\) into the formula.
When we take \(h + a\), we have \(5+4=9\) and \(y\) - coordinate \(k = 3\), so one vertex is \((9,3)\).
When we take \(h - a\), we have \(5 - 4=1\) and \(y\) - coordinate \(k = 3\), so the other vertex is \((1,3)\).

Answer:

Center point: \((5,3)\)
Vertices: \((1,3),(9,3)\)