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#