Find the Foci of the conic. $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 1$
A)Foci: $( 0 , \pm \sqrt { 48 } )$
B)Foci: $( \pm \sqrt { 8 } , 0 )$
C)Foci: $( 0 , \pm \sqrt { 8 } )$
D)Foci: $( \pm \sqrt { 48 } , 0 )$
E)Foci: $( \sqrt { 8 } , \sqrt { 48 } )$

Question

Find the Foci of the conic.  $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 1$ 
A)Foci: $( 0 , \pm \sqrt { 48 } )$
B)Foci: $( \pm \sqrt { 8 } , 0 )$
C)Foci: $( 0 , \pm \sqrt { 8 } )$
D)Foci: $( \pm \sqrt { 48 } , 0 )$
E)Foci: $( \sqrt { 8 } , \sqrt { 48 } )$

Quizplus · Accepted Answer

The given equation is of an ellipse in the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 16$ and $b^2 = 64$. Since $b^2 > a^2$, it's a vertically oriented ellipse. The foci are calculated using $c^2 = b^2 - a^2$, which gives $c^2 = 64 - 16 = 48$, so the foci are at $(0, \pm \sqrt{48})$.

Find the Foci of the Conic x216+y264=1\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 116x2​+64y2​=1

Find the Foci of the Conic $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 64 } = 1$