Question

the equation of the ellipse that has a center at $(4,2)$, a focus at $(0,2)$, and a vertex at $(-1,2)$, is
$\frac{(x - c)^2}{a^2}+\frac{(y - d)^2}{b^2}=1$
where
$a =$
$b =$
$c =$
$d =$

Explanation:

Step1: Determine C and D (center coordinates)

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

Step2: Find the value of A (distance from center to vertex)

The vertex is at \((- 1,2)\) and the center is at \((4,2)\). Since the y - coordinates of the center and the vertex are the same (\(y = 2\)), the ellipse is horizontal. The distance between the center \((4,2)\) and the vertex \((-1,2)\) is given by the formula for the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) which is \(d=\vert x_1 - x_2\vert\) (since \(y_1=y_2\)). So, \(A=\vert4-(-1)\vert=\vert4 + 1\vert = 5\).

Step3: Find the value of c (distance from center to focus)

The focus is at \((0,2)\) and the center is at \((4,2)\). Again, since the y - coordinates are the same, the distance between the center \((4,2)\) and the focus \((0,2)\) is \(c=\vert4 - 0\vert=4\).

Step4: Find the value of B (using the relationship \(c^{2}=A^{2}-B^{2}\))

For an ellipse, the relationship between \(A\) (semi - major axis), \(B\) (semi - minor axis) and \(c\) (distance from center to focus) is \(c^{2}=A^{2}-B^{2}\). We know \(A = 5\) and \(c = 4\). Substituting these values into the formula:

$$
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$$

Answer:

\(A = 5\), \(B=3\), \(C = 4\), \(D = 2\)