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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 20:43:53 IST
Accepted Answer [?]
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let S, S' r its foci n C its centre
S (x1,y1), S' (x2,y2) C (h,k)
SS'=2ae
(x1-x2)^2 + (y1-y2)^2=4a^2e^2
(x1+x2)^2 + (y1+y2)^2 -4(x1x2+y1y2) =4a^2e^2 = (c say)
C is mid point of S, S'
2h=x1+x2 and 2k=y1+y2
x1, x2 r perpendicular distances frm its foci on a tangent so r y1, y2
therefore by property of ellipse, x1x2=y1y2=b^2
now the eqn becomes
4h^2+4y^2-8b^2=c=4(a^2-b^2) (a^2e^2=a^2-b^2)
h^2+k^2=a^2+b^2
its a circle
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