How do you find the coordinates of the center, foci, the length of the major and minor axis given #3x^2+9y^2=27#?

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Answer 1 · 2016-12-29T07:20:33

The center is #(0,0)#
The foci are #F=(sqrt6,0)# and #F'=(-sqrt6,0)#
The length of the major axis is #=6#
The length of the minor axis is #=2sqrt3#

Let 's rearrange the equation

#3x^2+9y^2=27#

Dividing by #27#

#(3x^2)/27+(9y^2)/27=1#

#x^2/9+y^2/3=1#

#x^2/3^2+y^2/(sqrt3)^2=1#

We compare this to the general equation of the ellipse

#(x-h)^2/a^2+(y-k)^2/b^2=1#

#a=3# and #b=sqrt3#

The center is #(h.k)=(0,0)#

To calculate the foci, we need

#c=sqrt(a^2-b^2)=sqrt(9-3)=sqrt6#

The foci are F#=(h+c,k)=(sqrt6,0)#

and F'#=(h-c,k)=(-sqrt6,0)#

The length of the major axis is #=2a=6#

The length of the minor axis is #=2b=2sqrt3#

graph{3x^2+9y^2=27 [-5.546, 5.55, -2.773, 2.774]}