How do you find the coordinates of the center, foci, the length of the major and minor axis given #(x+8)^2/144+(y-2)^2/81=1#?

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Answer 1 · 2017-05-18T13:18:50

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#(x+8)^2/(12^2)+(y-2)^2/(9^2)=1#

We compare this equation to the standard equation of a horizontal ellipse

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

The center of the ellipse is #(h,k)=(-8,2)#

The length of the major axis is #=2a=2*12=24#

The length of the major axis is #=2b=2*9=18#

We need #c# to determine the foci

#c^2=a^2-b^2=144-81=53#

#c=+-sqrt53#

The foci are #F=(c,2)=(-8+sqrt53,2)# and

#F'=(-8-sqrt53,2)#
graph{((x+8)^2/144+(y-2)^2/81-1)((x+8)^2+(y-2)^2-0.1)=0 [-26.1, 5.95, -5.07, 10.95]}