Question

review for final examination
score: 39/50 answered: 39/50
question 40
given the ellipse \\(\frac{(x - 6)^2}{4} + \frac{(y - 3)^2}{16} = 1\\)
find the center point:
enter a point more..
list the four vertices/co-vertices:
question help: video 1 video 2

Explanation:

Step1: Recall ellipse standard form

The standard form of an ellipse centered at \((h,k)\) is \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (for vertical major axis) or \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (for horizontal major axis), where \((h,k)\) is the center.

Step2: Identify \(h\) and \(k\) from the given equation

Given the ellipse equation \(\frac{(x - 6)^2}{4}+\frac{(y - 3)^2}{16}=1\), comparing with the standard form \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\), we have \(h = 6\) and \(k=3\). So the center \((h,k)\) is \((6,3)\).

Answer:

\((6, 3)\)