Find the foci of the ellipse whose equation is given.
-$$36 ( \mathrm { x } - 1 ) ^ { 2 } + 25 ( \mathrm { y } - 3 ) ^ { 2 } = 900$$
A) foci at $( 1,3 - \sqrt { 11 } )$ and $( 1,3 + \sqrt { 11 } )$
B) foci at $( 3,1 - \sqrt { 11 } )$ and $( 3,1 + \sqrt { 11 } )$
C) foci at $( - 1,3 - \sqrt { 11 } )$ and $( - 1,3 + \sqrt { 11 } )$
D) foci at $( 2,3 - \sqrt { 11 } )$ and $( 2,3 + \sqrt { 11 } )$

Question

Quizplus · Accepted Answer

The given equation of the ellipse is in the form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $h=1$, $k=3$, $a^2=900/36=25$, and $b^2=900/25=36$. Since $b^2 > a^2$, it's a vertical ellipse. The foci are calculated using $c=\sqrt{b^2-a^2}=\sqrt{36-25}=\sqrt{11}$. The foci are at $(h, k \pm c) = (1, 3 \pm \sqrt{11})$.

Find the Foci of the Ellipse Whose Equation Is Given 36(x−1)2+25(y−3)2=90036 ( \mathrm { x } - 1 ) ^ { 2 } + 25 ( \mathrm { y } - 3 ) ^ { 2 } = 90036(x−1)2+25(y−3)2=900

Find the Foci of the Ellipse Whose Equation Is Given $36 ( \mathrm { x } - 1 ) ^ { 2 } + 25 ( \mathrm { y } - 3 ) ^ { 2 } = 900$