Find an equation for the ellipse described.
-Center at (3, 3); focus at (9, 3); vertex at (11, 3)
A) $\frac { ( x + 3 ) ^ { 2 } } { 36 } - \frac { ( y - 3 ) ^ { 2 } } { 22 } = 1$
B) $\frac { ( x - 3 ) ^ { 2 } } { 64 } + \frac { ( y - 3 ) ^ { 2 } } { 28 } = 1$
C) $\frac { ( x + 3 ) ^ { 2 } } { 64 } + \frac { ( y + 3 ) ^ { 2 } } { 28 } = 1$
D) $\frac { ( x - 3 ) ^ { 2 } } { 121 } + \frac { ( y + 3 ) ^ { 2 } } { 10 } = 2$

Question

Find an equation for the ellipse described.
-Center at (3, 3); focus at (9, 3); vertex at (11, 3) 
A) $\frac { ( x + 3 ) ^ { 2 } } { 36 } - \frac { ( y - 3 ) ^ { 2 } } { 22 } = 1$
B) $\frac { ( x - 3 ) ^ { 2 } } { 64 } + \frac { ( y - 3 ) ^ { 2 } } { 28 } = 1$
C) $\frac { ( x + 3 ) ^ { 2 } } { 64 } + \frac { ( y + 3 ) ^ { 2 } } { 28 } = 1$
D) $\frac { ( x - 3 ) ^ { 2 } } { 121 } + \frac { ( y + 3 ) ^ { 2 } } { 10 } = 2$

Quizplus · Accepted Answer

The center of the ellipse is at (3, 3), indicating that the ellipse is centered on these coordinates, so the equation will have the form $\frac{(x-3)^2}{a^2} + \frac{(y-3)^2}{b^2} = 1$, where $a$ and $b$ are the semi-major and semi-minor axes, respectively. The focus is at (9, 3), which is 6 units away from the center, indicating that the focal length $c = 6$. The vertex is at (11, 3), which is 8 units away from the center, indicating that the semi-major axis $a = 8$. Therefore, $a^2 = 64$. Since the ellipse is horizontal (major axis is along the x-axis), we use the formula $c^2 = a^2 - b^2$ to find $b^2$. Plugging in the values, we get $36 = 64 - b^2$, which gives $b^2 = 28$. Therefore, the correct equation is $\frac{(x-3)^2}{64} + \frac{(y-3)^2}{28} = 1$.

Find an Equation for the Ellipse Described (x+3)236−(y−3)222=1\frac { ( x + 3 ) ^ { 2 } } { 36 } - \frac { ( y - 3 ) ^ { 2 } } { 22 } = 136(x+3)2​−22(y−3)2​=1

Find an Equation for the Ellipse Described $\frac { ( x + 3 ) ^ { 2 } } { 36 } - \frac { ( y - 3 ) ^ { 2 } } { 22 } = 1$