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

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Answer 1 · 2018-06-06T17:30:55

The foci are;
$\pm 4, 0$ $+-4,0$

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

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

$\frac{x^{2}}{80 \times \frac{1}{4}} + \frac{y^{2}}{80 \times \frac{1}{20}} = 1$ $x^2/(80xx1/4)+y^2/(80xx1/20)=1$

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

Standard form of the ellipse

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

$a^{2} = 20 \to a = \sqrt{20} = 2 \sqrt{5}$ $a^2=20->a=sqrt(20)=2sqrt5$

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

Further, in conic sections related to ellipse

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

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

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

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

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

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

$e = \frac{2}{\sqrt{5}}$ $e=2/sqrt5$

The foci are,

$(\pm a e, 0)$ $(+-ae,0)$

$\pm 2 \sqrt{5} \times \frac{2}{\sqrt{5}}, 0$ $+-2sqrt5xx2/sqrt5,0$

$\pm 4, 0$ $+-4,0$

How do you find the foci for 4x^2 + 20y^2 = 804x2+20y2=804x^2 + 20y^2 = 80?