Find the equation of the ellipse with center at $( 8,9 )$, vertex at $( 16,9 )$, and passing through the intersection of $y = - 2 x + 27$ and $y = x + 3$.
A) $\frac { ( x - 8 ) ^ { 2 } } { 8 } + \frac { ( y - 9 ) ^ { 2 } } { 2 } = 1$
B) $\frac { ( x - 8 ) ^ { 2 } } { 49 } + \frac { ( y - 9 ) ^ { 2 } } { 9 } = 1$
C) $\frac { ( x - 8 ) ^ { 2 } } { 64 } + \frac { ( y - 9 ) ^ { 2 } } { 4 } = 1$
D) $\frac { ( x - 8 ) ^ { 2 } } { 49 } + \frac { ( y - 9 ) ^ { 2 } } { 4 } = 1$
E) $\frac { ( x - 8 ) ^ { 2 } } { 81 } + \frac { ( y - 9 ) ^ { 2 } } { 9 } = 1$

Question

Quizplus · Accepted Answer

The center of the ellipse is at (8,9), and one vertex is at (16,9), indicating the major axis is horizontal and the distance from the center to this vertex is 8 units, which means $a=8$ and $a^2=64$. To find the intersection of $y=-2x+27$ and $y=x+3$, set the equations equal: $-2x+27=x+3$, solving gives $x=8$, and substituting back gives $y=11$. This point lies on the ellipse, and using the center (8,9), the vertical distance to this point is 2 units, indicating $b=2$ and $b^2=4$. Thus, the equation is $\frac{(x-8)^2}{64}+\frac{(y-9)^2}{4}=1$.

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

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