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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Nov 2007 21:47:19 IST
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show that whatever be the values of theta,the line y=(x-11) cos theta - cos 3theta is a normal to y2=16x
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28 Nov 2007 21:50:19 IST
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apply the same concept here as in ur previous question...eqn of normal
u'll get the answer
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Rahul Dey
Dept. of Electronics & Electrical
Communication Engineering,
IIT Kharagpur
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28 Nov 2007 21:57:31 IST
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tried but not getting it,pleaseeeeeeeeeeeee help
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28 Nov 2007 22:55:52 IST
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kindly somebody help
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y = mx -2am - am^3 { for parabola y^2=4ax}
so here eqn of noormal bcomes
y= mx-8m-4m^3 ..........(1)
the other eqn is y = xcos(theta) -11cos(theta)-cos3(theta)
y= xcos(theta) - 11cos(theta)-(4cos^3 (theta) - 3cos(theta))
y = x cos(theta) - 8cos(theta) -4cos^3(theta) .......(2)
now compare 1&2
see m is slope = tanh { h be any angle}
now on
-1<=cos(theta)<=1
always
now frm 1 we see that slope is tanh, frm 2 is cos(theta)
since both eqn are identical
we must have tanh=cos(theta)
now see here that range of cos(theta) is restricted but not of tanh
so
whatever b the value of cos(theta) there will be a value of 'h'
such that
tanh=cos(theta)
hence the given equation will always be a normal
for any theta
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28 Nov 2007 22:58:11 IST
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let theta = w
its the same thing
y+xt=2at+at^3
dis is
y=(x-11)cosw-cos3w
y-coswx=-11cosw-(4cos^3w-3cosw)=-8cosw-4cos^3w
again on comparing, t=-cosw, here a=4
so the same thing
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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
Rahul Dey
Dept. of Electronics & Electrical
Communication Engineering,
IIT Kharagpur
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28 Nov 2007 23:06:55 IST
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thnxs a lot
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