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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 10:21:16 IST
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the line lx + my +n = 0 is a normal to the ellipse x2/a2 + y2/b2 = 1 if (a) a2/l2 - b2/m2 = (a2 - b2)2 / n2 (b) a2/m2 + b2/l2 = (a2 - b2)2 / n2 (c) a2/l2 + b2/m2 = (a2 - b2)2 / n2 (d) none of these
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29 Nov 2007 10:46:25 IST
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eqn of normal to ellipse is y=Mx+rt(a^2M^2+b^2) (M is slope of normal)
in this line, M=-l/m
so equate rt(a^2l^2/m^2 + b^2)=-n/l n get the answer
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29 Nov 2007 10:57:41 IST
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hey isnt that the condition fr line to be tangent of ellipse?? i think the eqn to normal given as a^2 x/ x1 - b^2 y/ y1 = a^2- b^2 wud help cuz by ur method 'm not gettin a a^2- b^2 term as needed by the options.. here x1 n y1 lie on the ellipse..but how do u eliminate them both is the ques..
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oops sory!buzd up
now c
any point on ellipse will be (acosw, bsinw) (w is any angle)
normal at point w is
asecwx-bcosecwy=a^2-b^2
now compare
l/asecw=-m/bcosecw=-n/(a^2-b^2)
u get cosw=-an/l(a^2-b^2) n sinw=bn/m(a^2-b^2)
apply cos^2w+sin^2w=1
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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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29 Nov 2007 11:54:48 IST
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nudge me if u have any more doubts
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- Edward Everett Hale
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29 Nov 2007 11:55:33 IST
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thnks again..ans is right..opt c..
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29 Nov 2007 11:56:44 IST
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sure thing..right now i stil have few doubts floatin in this forum..y dont u try those..
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