Question

chapter 10 review
score: 75/100 answered: 8/10
question 10
given the ellipse \\( \frac{(x - 3)^2}{25}+\frac{(y - 2)^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 equation

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 \(\frac{(x - 3)^2}{25}+\frac{(y - 2)^2}{16}=1\), by comparing with the standard form, we can directly get the center.

Step2: Find the center

Since \(h = 3\) and \(k=2\), the center of the ellipse is \((3,2)\).

Step3: Determine the value of \(a\)

We know that \(a^2 = 25\), so \(a=\sqrt{25}=5\).

Step4: Find the vertices

The vertices of an ellipse of the form \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) are \((h\pm a,k)\).
Substituting \(h = 3\), \(a = 5\), and \(k = 2\) into the formula, we get the vertices:
When \(x=h + a\), \(x=3 + 5=8\), \(y = 2\), so one vertex is \((8,2)\).
When \(x=h - a\), \(x=3-5=-2\), \(y = 2\), so the other vertex is \((-2,2)\).

Answer:

Center point: \((3,2)\)
Vertices: \((-2,2),(8,2)\)