# How do you find the foci for 4x^2 + 20y^2 = 80?

The foci are;
$\pm 4 , 0$

#### Explanation:

$4 {x}^{2} + 20 {y}^{2} = 80$

${x}^{2} / \left(\frac{1}{4}\right) + {y}^{2} / \left(\frac{1}{20}\right) = 80$

${x}^{2} / \left(80 \times \frac{1}{4}\right) + {y}^{2} / \left(80 \times \frac{1}{20}\right) = 1$

${x}^{2} / 20 + {y}^{2} / 4 = 1$

Standard form of the ellipse

${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$

${a}^{2} = 20 \to a = \sqrt{20} = 2 \sqrt{5}$

${b}^{2} = 4 \to b = 2$

Further, in conic sections related to ellipse

${b}^{2} = {a}^{2} \left(1 - {e}^{2}\right)$

$4 = 20 \left(1 - {e}^{2}\right)$

$\frac{4}{20} = 1 - {e}^{2}$

$\frac{1}{5} = 1 - {e}^{2}$

${e}^{2} = 1 - \frac{1}{5}$

${e}^{2} = \frac{4}{5}$

$e = \frac{2}{\sqrt{5}}$

The foci are,

$\left(\pm a e , 0\right)$

$\pm 2 \sqrt{5} \times \frac{2}{\sqrt{5}} , 0$

$\pm 4 , 0$