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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Apr 2007 00:49:37 IST
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for any parabola what is the condition for a point such that three normals can be drawn from it to the parabola?
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4 Apr 2007 11:29:33 IST
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from every point on the parabola itself three normals can be drawn.
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4 Apr 2007 11:37:35 IST
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the pt should lie inside parabola not on parabola.
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4 Apr 2007 16:47:23 IST
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if the equation of the parabola is y2=4ax, then the equation of the normals in terms of the slope form is given by y=mx-2am-am3, for a point ,say (  ,  ), the equation becomes =m -2am-am 3=> am 3 + m(2a- ) + =0 for 3 normals to be possible, this equation should have 3 real roots , for that there are 2 conditions 1) >2a 2) f( ')f( ')<0 where ' & ' are the roots of the derivative of the equation-> am 3 + m(2a- ) + =0 f( ') & f( ') denote function. f(m)=am 3 + m(2a- ) + =0 i hope this helps...
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4 Apr 2007 19:02:17 IST
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Arnnie is correct... and if you are given specific values of the point's cordinate and 'a' of the parabola.. the the eqn. in m should have distinct roots.... that will be the sufficient condition.
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Sudeep Kumar
(B tech, IITd)
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4 Apr 2007 19:46:59 IST
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if the point lies on the axis then the x coordinate should be greater than 2a ifor y2=4ax.....put (x,0) in eqn of normal and check it out
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4 Apr 2007 20:05:04 IST
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the correct answer is that th x co-ordinate of the point should be greater than 2a
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4 Apr 2007 21:38:36 IST
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Arnnie could you please explain point no. 2????
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5 Apr 2007 10:55:56 IST
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Anjishnu said
Arnnie could you please explain point no. 2???? yea, sure man!! the equation am3+m(2a-)+=0 can be written as a function of "m". so it becomes, f(m)=am3+m(2a-)+=0 when you differentiate f(m) it become f''(m)=3am2+2a-=0, let this equation have roots '' & ''. when we put the values '' & '' in the 1st equation i.e. in f(m), we get f('') & f(''). so, the condition is that f('') * f('') <0. i hope that explains....
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6 Apr 2007 19:50:15 IST
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m1+ m2+ m3=0
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