Question

question 2
given the ellipse $\frac{(x - 4)^2}{25}+\frac{(y - 6)^2}{4}=1$,
find the center point:
list the vertices (separated by a comma):
question help: video written example

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), where \((h,k)\) is the center of the ellipse.
For the given ellipse \(\frac{(x - 4)^2}{25}+\frac{(y - 6)^2}{4}=1\), by comparing with the standard form, we can directly get the center \((h,k)\).
Here \(h = 4\) and \(k=6\).

Step2: Find the vertices

Since \(a^2 = 25\), then \(a = 5\) (because \(a>0\)). The major axis is parallel to the \(x\) - axis (since \(a^2>b^2\), here \(a^2 = 25\) and \(b^2=4\)).
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 \(h = 4\), \(k = 6\) and \(a = 5\) into the formula for vertices.
The vertices are \((4 + 5,6)\) and \((4-5,6)\)

Answer:

Center: \((4,6)\)
Vertices: \((9,6),( - 1,6)\)