Find the foci of the ellipse whose equation is given.
-$$\frac { ( x - 3 ) ^ { 2 } } { 36 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 1$$
A) foci at $( 3 + 2 \sqrt { 5 } , 1 )$ and $( 3 - 2 \sqrt { 5 } , 1 )$
B) foci at $( 1 + 2 \sqrt { 5 } , 3 )$ and $( 1 - 2 \sqrt { 5 } , 3 )$
C) foci at $( - 2 \sqrt { 5 } , 1 )$ and $( 2 \sqrt { 5 } , 1 )$
D) foci at $( 3 + 2 \sqrt { 5 } , 3 )$ and $( 3 - 2 \sqrt { 5 } , 3 )$

Question

Quizplus · Accepted Answer

The given ellipse equation is in the form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $a^2 = 36$, $b^2 = 16$, $h = 3$, and $k = 1$. Since $a^2 > b^2$, the ellipse is horizontal, and the foci are located along the x-axis from the center. The distance of each focus from the center is $c = \sqrt{a^2 - b^2} = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5}$. Therefore, the foci are at $(3 \pm 2\sqrt{5}, 1)$.

Find the Foci of the Ellipse Whose Equation Is Given (x−3)236+(y−1)216=1\frac { ( x - 3 ) ^ { 2 } } { 36 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 136(x−3)2​+16(y−1)2​=1

Find the Foci of the Ellipse Whose Equation Is Given $\frac { ( x - 3 ) ^ { 2 } } { 36 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 1$