The point (10,26) is a focus of a non-degenerate ellipse tangent to the positive x and y axes. The locus of the center of the ellipse lies along a line. Find the equation of this line.
The easiest ellipses have either horizontal and vertical axes. The ellipses will have two different centers and these points should determine the line sought.
The equation of an ellipse tangent to x and y axes is
(x-h)^2/h^2 + (y-k)^2/k^2 = 1
Case 1: Horizontal h>26
Focus =(h+/-c,k) a^2-b^2=c^2
(x-h)^2/h^2 + (y-26)^2/26^2 =1
c=h-10
h^2-26^2=(h-10)^2
h=194/5 = 38.8
Center point = (194/5,26)
Case 2: Vertical h<10
Focus =(h,k+/-c) b^2-a^2=c^2
(x-10)^2/10^2 + (y-k)^2/k^^ =1
c=k-26
k^2-10^2=(k-26)^2
k=194/13 = 14.923...
Center point = (10,194/13)
The center points determine the line y=(5/13)x+(144/13)
IF the locus is a line then this is the line.
I can imagine taking the red ellipse and rotating it the the blue one with the focus (10,26) remaining fixed. It's believable that the center point stays on the line since the ellipse will have to get bigger.
https://www.desmos.com/calculator/mef4xzuhvf
I have not proven this and I don't know if I will.
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Posted by Jer
on 2024-12-15 13:35:54 |
Answer with incomplete proof
