Question

the equation of the ellipse that has a center at $(9,4)$, a focus at $(5,4)$, and a vertex at $(4,4)$, is
$\frac{(x - c)^2}{a^2}+\frac{(y - d)^2}{b^2}=1$
where
$a =$
$b =$
$c =$
$d =$

Explanation:

Step1: Determine the values of \(C\) and \(D\)

The center of the ellipse is \((h,k)=(9,4)\). In the standard form \(\frac{(x - C)^{2}}{A^{2}}+\frac{(y - D)^{2}}{B^{2}} = 1\), we have \(C = 9\) and \(D=4\).

Step2: Calculate the value of \(A\)

The distance from the center \((9,4)\) to a vertex \((4,4)\) is given by \(A=\vert h - x_{vertex}\vert\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\), since \(y_2=y_1 = 4\), \(A=\vert9 - 4\vert=5\).

Step3: Calculate the value of \(c\)

The distance from the center \((9,4)\) to a focus \((5,4)\) is \(c=\vert h - x_{focus}\vert\). Since \(y_2=y_1 = 4\), \(c=\vert9 - 5\vert = 4\).

Step4: Calculate the value of \(B\)

For an ellipse, the relationship \(c^{2}=A^{2}-B^{2}\) holds. We know \(A = 5\) and \(c = 4\). Substituting into the formula \(B^{2}=A^{2}-c^{2}\), we get \(B^{2}=25 - 16=9\), so \(B = 3\).

Answer:

\(A = 5\)
\(B=3\)
\(C = 9\)
\(D = 4\)