Question

the equation of the ellipse that has a center at $(4,1)$, a focus at $(1,1)$, and a vertex at $(9,1)$, is
$\frac{(x - c)^2}{a^2}+\frac{(y - d)^2}{b^2}=1$
where
$a =$
$b =$
$c =$
$d =$
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Explanation:

Step1: Determine C and D (center coordinates)

The center of the ellipse is given as \((4, 1)\). In the standard form \(\frac{(x - C)^2}{A^2} + \frac{(y - D)^2}{B^2} = 1\), the center is \((C, D)\). So, \(C = 4\) and \(D = 1\).

Step2: Determine A (distance from center to vertex)

The vertex is at \((9, 1)\) and the center is at \((4, 1)\). Since they have the same \(y\)-coordinate, the distance \(A\) is the horizontal distance between them. Using the distance formula (or just subtracting \(x\)-coordinates since \(y\) is constant): \(A=\vert9 - 4\vert= 5\).

Step3: Determine c (distance from center to focus)

The focus is at \((1, 1)\) and the center is at \((4, 1)\). Again, same \(y\)-coordinate, so the distance \(c\) (distance from center to focus) is \(\vert1 - 4\vert = 3\).

Step4: Determine B using the ellipse relationship \(c^2=A^2 - B^2\)

We know the relationship for an ellipse is \(c^2=A^2 - B^2\). We have \(c = 3\) and \(A = 5\). Plugging in: \(3^2=5^2 - B^2\)
\(9 = 25 - B^2\)
Solving for \(B^2\): \(B^2=25 - 9=16\), so \(B = \sqrt{16}=4\) (we take the positive root since \(B\) represents a length).

Answer:

\(A = 5\), \(B = 4\), \(C = 4\), \(D = 1\)