How do I find the foci of an ellipse if its equation is $\frac{x^{2}}{16} + \frac{y^{2}}{36} = 1$ $x^2/16+y^2/36=1$ ?

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How do I find the foci of an ellipse if its equation is $\frac{x^{2}}{16} + \frac{y^{2}}{36} = 1$ $x^2/16+y^2/36=1$ ?

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Answer 1 · 2014-10-04T21:30:59

$\frac{x^{2}}{16} + \frac{y^{2}}{36} = 1$ $x^2/16+y^2/36=1$

$a^{2} = 36$ $a^2=36$
$b^{2} = 16$ $b^2=16$

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

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

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

$c^{2} = a^{2} - b^{2}$ $c^2=a^2-b^2$
$c^{2} = 36 - 16$ $c^2=36-16$
$c^{2} = 20$ $c^2=20$
$c = \sqrt{2 \cdot 2 \cdot 5}$ $c=sqrt(2*2*5)$
$c = 2 \sqrt{5}$ $c=2sqrt(5)$

The coordinates for the foci are $(0, \pm c) \to (0, \pm 2 \sqrt{5})$ $(0,+-c) -> (0,+-2sqrt(5))$

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How do I find the foci of an ellipse if its equation is $\frac{x^{2}}{16} + \frac{y^{2}}{36} = 1$ $x^2/16+y^2/36=1$ ?

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