Question

review for final examination
score: 30/50 answered: 30/50
question 14
the equation of the ellipse that has a center at (6,3), a focus at (3,3), and a vertex at (11,3), is
\frac{(x - c)^2}{a^2} + \frac{(y - d)^2}{b^2} = 1
where
a =

b =

c =

d =

Explanation:

Step1: Find \(A\)

The distance from the center \((6,3)\) to a vertex \((11,3)\) is \(A\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), here \(y_1 = y_2 = 3\), so \(A=\vert11 - 6\vert=5\)

Step2: Find \(C\) and \(D\)

The center of the ellipse is \((h,k)=(6,3)\). In the standard form \(\frac{(x - h)^2}{A^2}+\frac{(y - k)^2}{B^2}=1\), we have \(C = h=6\) and \(D = k = 3\)

Step3: Find \(c\) and then \(B\)

The distance from the center \((6,3)\) to a focus \((3,3)\) is \(c\). Using the distance formula (since \(y_1=y_2 = 3\)), \(c=\vert6 - 3\vert=3\)
For an ellipse, \(c^2=A^2 - B^2\). Substituting \(A = 5\) and \(c = 3\), we get \(3^2=5^2 - B^2\)
\(B^2=25 - 9=16\), so \(B = 4\)

Answer:

\(A = 5\)
\(B = 4\)
\(C = 6\)
\(D = 3\)