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

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Answer 1 · 2018-06-06T17:30:55

The foci are;
$+-4,0$

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

$x^2/(1/4)+y^2/(1/20)=80$

$x^2/(80xx1/4)+y^2/(80xx1/20)=1$

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

Standard form of the ellipse

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

$a^2=20->a=sqrt(20)=2sqrt5$

$b^2=4->b=2$

Further, in conic sections related to ellipse

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

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

$4/20=1-e^2$

$1/5=1-e^2$

$e^2=1-1/5$

$e^2=4/5$

$e=2/sqrt5$

The foci are,

$(+-ae,0)$

$+-2sqrt5xx2/sqrt5,0$

$+-4,0$

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