Question

given the ellipse $\frac{(x - 4)^2}{25}+\frac{(y - 6)^2}{9}=1$,
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 centered at \((h, k)\) is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (when \(a>b\), the major axis is horizontal) or \(\frac{(y - k)^2}{a^2}+\frac{(x - h)^2}{b^2}=1\) (when \(a>b\), the major axis is vertical). Here, our equation is \(\frac{(x - 4)^2}{25}+\frac{(y - 6)^2}{9}=1\), which is in the form \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) with \(h = 4\), \(k=6\), \(a^2 = 25\) (so \(a = 5\)) and \(b^2=9\) (so \(b = 3\)).

Step2: Find the center

For the ellipse \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), the center is \((h,k)\). Comparing with \(\frac{(x - 4)^2}{25}+\frac{(y - 6)^2}{9}=1\), we have \(h = 4\) and \(k = 6\). So the center is \((4,6)\).

Step3: Find the vertices

Since the major axis is horizontal (because \(a^2=25\) is under the \((x - h)^2\) term and \(a>b\) as \(5>3\)), the vertices are located at \((h\pm a,k)\). We know \(h = 4\), \(a = 5\) and \(k = 6\). So the vertices are \((4 + 5,6)=(9,6)\) and \((4-5,6)=(- 1,6)\).

Answer:

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