Find the foci of the ellipse whose equation is given.
-$$\frac { ( x - 3 ) ^ { 2 } } { 25 } + \frac { ( y + 2 ) ^ { 2 } } { 36 } = 1$$
A) foci at $( 3 , - 2 - \sqrt { 11 } )$ and $( 3 , - 2 + \sqrt { 11 } )$
B) foci at $( - 2,3 - \sqrt { 11 } )$ and $( - 2,3 + \sqrt { 11 } )$
C) foci at $( - 3 , - 2 - \sqrt { 11 } )$ and $( - 3 , - 2 + \sqrt { 11 } )$
D) foci at $( 4 , - 2 - \sqrt { 11 } )$ and $( 4 , - 2 + \sqrt { 11 } )$

Question

Quizplus · Accepted Answer

The given equation is in the form of an ellipse centered at $(h, k)$ with the form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $a^2$ and $b^2$ are the denominators of the $x$ and $y$ terms, respectively. For this ellipse, $h = 3$, $k = -2$, $a^2 = 25$, and $b^2 = 36$. Since $b^2 > a^2$, this is a vertically oriented ellipse. The distance of the foci from the center along the major axis is given by $c$, where $c^2 = b^2 - a^2$. Thus, $c^2 = 36 - 25 = 11$, so $c = \sqrt{11}$. The foci are therefore at $(h, k \pm c) = (3, -2 \pm \sqrt{11})$.

Find the Foci of the Ellipse Whose Equation Is Given (x−3)225+(y+2)236=1\frac { ( x - 3 ) ^ { 2 } } { 25 } + \frac { ( y + 2 ) ^ { 2 } } { 36 } = 125(x−3)2​+36(y+2)2​=1

Find the Foci of the Ellipse Whose Equation Is Given $\frac { ( x - 3 ) ^ { 2 } } { 25 } + \frac { ( y + 2 ) ^ { 2 } } { 36 } = 1$