How do I find the foci of an ellipse if its equation is $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ $x^2/16+y^2/9=1$ ?

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Answer 1 · 2014-10-04T21:28:01

$\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ $x^2/16+y^2/9=1$

$a^{2} = 16$ $a^2=16$
$b^{2} = 9$ $b^2=9$

The center is $(0, 0)$ $(0,0)$ .

The x-axis is the major axis. We know this because the $x^{2}$ $x^2$ term has the larger denominator.

We find the foci using the following equation and solving for $c$ $c$ .

$c^{2} = a^{2} - b^{2}$ $c^2=a^2-b^2$
$c^{2} = 16 - 9$ $c^2=16-9$
$c^{2} = 7$ $c^2=7$
$c = \sqrt{7}$ $c=sqrt(7)$

The coordinates for the foci are $(\pm c, 0) \to (\pm \sqrt{7}, 0)$ $(+-c,0) -> (+-sqrt(7),0)$

How do I find the foci of an ellipse if its equation is x216+y29=1x^2/16+y^2/9=1?