Find the standard form of the equation of the ellipse with the given characteristics.
center: $( - 8 , - 5 ) \quad a = 5 c \quad$ foci: $( - 11 , - 5 ) , ( - 5 , - 5 )$
A) $\frac { ( x - 8 ) ^ { 2 } } { 216 } + \frac { ( y - 5 ) ^ { 2 } } { 225 } = 1$
B) $\frac { ( x + 8 ) ^ { 2 } } { 216 } + \frac { ( y + 5 ) ^ { 2 } } { 225 } = 1$
C) $\frac { ( x + 8 ) ^ { 2 } } { 225 } + \frac { ( y + 5 ) ^ { 2 } } { 216 } = 1$
D) $\frac { ( x + 5 ) ^ { 2 } } { 225 } + \frac { ( y + 8 ) ^ { 2 } } { 216 } = 1$
E) $\frac { ( x - 5 ) ^ { 2 } } { 225 } + \frac { ( y - 8 ) ^ { 2 } } { 216 } = 1$

Question

Quizplus · Accepted Answer

First, we determine that the center of the ellipse is at $(-8,-5)$, which gives us the translation factor for both the $x$ and $y$ terms of the equation. Thus, the equation of the ellipse is of the form: $$\frac{(x+8)^2}{a^2}+\frac{(y+5)^2}{b^2}=1$$  Next, we use the fact that $a=5c$, and since the foci are at $(-11,-5)$ and $(-5,-5)$, the distance between the two foci is $|(-11)-(-5)|=6=2c$. Therefore, $c=3$, and $a=5c=15$, allowing us to solve for $b$ using the Pythagorean Theorem: $$b^2=a^2-c^2=15^2-3^2=216$$  Thus, the standard form of the equation is $$\frac{(x+8)^2}{225}+\frac{(y+5)^2}{216}=1$$ which corresponds to answer choice $\boxed{	extbf{(C)}}$.

Find the Standard Form of the Equation of the Ellipse (−8,−5)a=5c( - 8 , - 5 ) \quad a = 5 c \quad(−8,−5)a=5c

Find the Standard Form of the Equation of the Ellipse $( - 8 , - 5 ) \quad a = 5 c \quad$