How do you write an equation of an ellipse in standard form given foci at (- 6, 5) and (- 6, - 7) and whose major axis has length 26?

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Answer 1 · 2017-11-02T15:39:13

The equation is #(y+1)^2/169+(x+6)^2/133=1#

The major axis is vertical

The foci are

#F=(-6,5)=(x_0,y_0+c)#

#F'=(-6,-7)=(x_0,y_0-c)#

Therefore,

#x_0=-6#

#y_0=-1#

The center of the ellipse is #C=(x_0,y_0)=(-6,-1)#

Also,

#2a=26#, #=>#, #a=13#

#FF'=2c=12#, #=>#, #c=6#

Therefore,

#b^2=a^2-c^2=13^2-6^2=169-36=133#

#b=sqrt133#

The equation of the ellipse is

#(y-y_0)^2/a^2+(x-x_0)^2/b^2=1#

#(y+1)^2/169+(x+6)^2/133=1#

graph{((y+1)^2/169+(x+6)^2/133-1)(y-1000(x+6))=0 [-36.9, 14.42, -12.38, 13.3]}