How do I write an equation in general form of an ellipse with foci at (5,2) and (5, -10) and another point at (7,15)?

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Answer 1 · 2018-07-26T09:44:16

The equation is #(y+4)^2/365.45+(x-5)^2/329.45=1#

The center of the ellipse is

#C=((5+5)/2,(2-10)/2)=(5,-4)#

This is an ellipse with vertical major axis

The equation of the ellipse is

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

#(y+4)^2/a^2+(x-5)^2/b^2=1#

As the point #(7,15)# lies on the ellipse

#(15-(-4))^2/a^2+(7-5)^2/b^2=1#

#361/a^2+4/b^2=1#.............................#(1)#

Also

#a^2=c^2+b^2#

#a^2=36+b^2#...................#(2)#

Solving for #a^2# and #b^2# in equations #(1)# and #(2)#

#361/(b^2+36)+4/b^2=1#

#361(b^2)+4(b^2+36)=(b^2+36)(b^2)#

#361b^2+4b^2+144=b^4+36b^2#

#b^4-329b^2-144=0#

#b^2=(329+-sqrt(329^2+4*144))/2#

#=(329+329.9)/2#

#=329.45#

And

#a^2=36+329.45=365.45#

The equation is

#(y+4)^2/365.45+(x-5)^2/329.45=1#

graph{(y+4)^2/365.45+(x-5)^2/329.45-1=0 [-46.22, 46.24, -23.13, 23.13]}