Find the equation of the ellipse with center at $( 8,9 )$, vertex at $( 16,9 )$, and passing through the intersection of $y = - 2 x + 32$ and $y = x + 8$.
A) $\quad \frac { ( x - 8 ) ^ { 2 } } { 8 } + \frac { ( y - 9 ) ^ { 2 } } { 7 } = 1$
B) $\quad \frac { ( x - 8 ) ^ { 2 } } { 64 } + \frac { ( y - 9 ) ^ { 2 } } { 49 } = 1$
C) $\quad \frac { ( x - 8 ) ^ { 2 } } { 49 } + \frac { ( y - 9 ) ^ { 2 } } { 49 } = 1$
D) $\quad \frac { ( x - 8 ) ^ { 2 } } { 81 } + \frac { ( y - 9 ) ^ { 2 } } { 64 } = 1$
E)
$$\frac { ( x - 8 ) ^ { 2 } } { 49 } + \frac { ( y - 9 ) ^ { 2 } } { 64 } = 1$$

Question

Quizplus · Accepted Answer

The center of the ellipse is at $(8,9)$, and one vertex is at $(16,9)$, indicating that the major axis is horizontal and the distance from the center to this vertex is 8 units, which is the semi-major axis length, $a=8$. Thus, $a^2=64$. To find $b^2$, we first find the intersection of $y=-2x+32$ and $y=x+8$ by solving the system of equations, which gives us the point $(8,16)$. This point lies on the ellipse and provides the length of the semi-minor axis, $b$, when considering the distance in the $y$-direction from the center $(8,9)$ to $(8,16)$, which is 7 units, so $b=7$ and $b^2=49$. Therefore, the equation of the ellipse is $\frac{(x-8)^2}{64} + \frac{(y-9)^2}{49} = 1$.

Find the Equation of the Ellipse with Center At (8,9)( 8,9 )(8,9)

Find the Equation of the Ellipse with Center At $( 8,9 )$