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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 10:06:53 IST
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An ellipse slides between two perpendicular straight lines. the locus of its centre is: (a) straight line (b) circle (c) ellipse (d) hyperbola
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 20:13:26 IST
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any clue anyone??
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 20:37:11 IST
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its a circle wait i'm giving a complete soln
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
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I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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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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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 21:08:52 IST
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brilliant..thnks dude..
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