Question

given the ellipse \\( \frac{(x - 6)^{2}}{16}+\frac{(y - 3)^{2}}{25}=1 \\).
find the center point:
list the four vertices:

Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse is \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (where \(a>b\)), and the center of the ellipse is \((h,k)\).

Step2: Identify \(h\) and \(k\)

For the given ellipse \(\frac{(x - 6)^2}{16}+\frac{(y - 3)^2}{25}=1\), we have \(h = 6\) and \(k=3\).

Answer:

The center point is \((6,3)\)

For the vertices:

Step1: Recall the formula for vertices

Since \(a^2 = 25\), then \(a = 5\). The vertices of the ellipse \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) are \((h,k\pm a)\)

Step2: Calculate the vertices

Substitute \(h = 6\), \(k = 3\) and \(a = 5\) into the formula.
The vertices are \((6,3 + 5)=(6,8)\) and \((6,3-5)=(6,-2)\)

Since \(b^2=16\), then \(b = 4\). The co - vertices (which are also considered in the "four vertices" in the context of an ellipse's important points) are \((h\pm b,k)\)
Substitute \(h = 6\), \(k = 3\) and \(b = 4\) into the formula.
The co - vertices are \((6+4,3)=(10,3)\) and \((6 - 4,3)=(2,3)\)