Question

assignment 10.1: the ellipse
score: 40/100 answered: 2/5
question 3
given the ellipse \\( \frac{(x - 6)^{2}}{10}+\frac{(y - 3)^{2}}{25}=1 \\),
find the center point:
list the four vertices:
question help: video 1 video 2 post to forum
submit question

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 major axis is vertical). The center of the ellipse is \((h,k)\).
For the given ellipse \(\frac{(x - 0)^2}{10}+\frac{(y - 3)^2}{25}=1\), comparing with the standard form \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\), we have \(h = 0\) and \(k=3\).

Step2: Find the vertices

Since the major axis is vertical (because \(a^2 = 25\) and \(b^2=10\), so \(a = 5\) and \(b=\sqrt{10}\)), the vertices of the ellipse \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) are given by \((h,k\pm a)\) and \((h\pm b,k)\) (but for the standard - form we are using, the vertices along the major axis are \((h,k\pm a)\)).
Substituting \(h = 0\), \(k = 3\), and \(a = 5\) into \((h,k\pm a)\), we get \((0,3 + 5)=(0,8)\) and \((0,3-5)=(0,-2)\). Also, for the minor - axis vertices (using \(b=\sqrt{10}\)), we have \((0+\sqrt{10},3)=(\sqrt{10},3)\) and \((0-\sqrt{10},3)=(-\sqrt{10},3)\)

Answer:

Center point: \((0,3)\)
Vertices: \((0,8)\), \((0,-2)\), \((\sqrt{10},3)\), \((-\sqrt{10},3)\)