Question

given the ellipse $\frac{(x - 3)^2}{4}+\frac{(y - 6)^2}{25}=1$,
find the center point:
list the vertices:
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 \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (when \(a>b\), major axis is vertical) or \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (when \(a>b\), major axis is horizontal), where \((h,k)\) is the center, \(a\) is the semi - major axis length and \(b\) is the semi - minor axis length.

For the given ellipse equation \(\frac{(x - 3)^2}{4}+\frac{(y - 6)^2}{25}=1\), we can compare it with the standard form \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (since \(25>4\), the major axis is vertical).

Step2: Find the center of the ellipse

By comparing \(\frac{(x - 3)^2}{4}+\frac{(y - 6)^2}{25}=1\) with \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\), we can see that \(h = 3\) and \(k=6\). So the center of the ellipse \((h,k)\) is \((3,6)\).

Step3: Find the vertices of the ellipse

For an ellipse with a vertical major axis (since \(a^{2}=25\), so \(a = 5\) and \(b^{2}=4\), so \(b = 2\)) and center \((h,k)=(3,6)\), the vertices are located at \((h,k\pm a)\).

We know that \(h = 3\), \(k = 6\) and \(a=\sqrt{25}=5\).

So the vertices are \((3,6 + 5)=(3,11)\) and \((3,6-5)=(3,1)\).

Answer:

• Center point: \((3,6)\)
• Vertices: \((3,1)\) and \((3,11)\)