Question

assignment 8.1 the ellipse
score: 2/9 answered: 2/9
question 3
given the ellipse \\( \frac { ( x - 4 ) ^ { 2 } } { 9 } + \frac { ( y - 5 ) ^ { 2 } } { 4 } = 1 \\)
find the center point:
list the four vertices, as ordered pairs, separated by commas:
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Explanation:

Step1: Recall the standard form of an ellipse

The standard form of an ellipse centered at \((h, k)\) is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) (for a horizontal major axis) or \(\frac{(y - k)^2}{a^2}+\frac{(x - h)^2}{b^2}=1\) (for a vertical major axis), where \(a>b>0\).

Step2: Identify the center \((h, k)\) from the given equation

Given the ellipse equation \(\frac{(x - 4)^2}{9}+\frac{(y - 5)^2}{4}=1\), comparing with the standard form \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), we can see that \(h = 4\) and \(k=5\). So the center of the ellipse is \((4, 5)\).

Step3: Determine the values of \(a\) and \(b\)

From the equation \(\frac{(x - 4)^2}{9}+\frac{(y - 5)^2}{4}=1\), we have \(a^{2}=9\) and \(b^{2}=4\). Taking the square roots (since \(a,b>0\)), we get \(a = 3\) and \(b = 2\).

Step4: Find the vertices

Since \(a^{2}=9\) (under the \(x\)-term) is greater than \(b^{2}=4\) (under the \(y\)-term), the major axis is horizontal.

• The vertices along the major axis (horizontal) are \((h\pm a,k)\). Substituting \(h = 4\), \(a=3\), and \(k = 5\), we get \((4 + 3,5)=(7,5)\) and \((4-3,5)=(1,5)\).
• The vertices along the minor axis (vertical) are \((h,k\pm b)\). Substituting \(h = 4\), \(b = 2\), and \(k=5\), we get \((4,5 + 2)=(4,7)\) and \((4,5-2)=(4,3)\).

Answer:

Center point: \((4, 5)\)
Four vertices: \((7, 5),(1, 5),(4, 7),(4, 3)\)