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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Jan 2008 18:10:04 IST
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Write the equation of the common normal to y^2=4ax and x^2=4ay!!
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28 Jan 2008 18:25:55 IST
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it is y=2x-12a, y=-x+3a
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28 Jan 2008 18:27:06 IST
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hey man, can u show the working plz.. Ill give u good rating if it is good.
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28 Jan 2008 18:34:49 IST
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the common normal of both parabolas shud be normal to y=x also.
so its slope is -1.
now juss, y=mx-2am-am^3, put m=-1
x+y=3a
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28 Jan 2008 18:43:46 IST
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hey rhd92781 could u show ur steps i did'nt understand ur concept man!!!!!!!!
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28 Jan 2008 18:45:07 IST
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hey yaar, you did'nt write anything??
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28 Jan 2008 18:48:13 IST
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hey raulrag,why dont' you show your steps...man,,?
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28 Jan 2008 18:58:53 IST
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SEE man , I hav purposly changed one step in my working .. keep seeing my steps, i'll notify it when i change it Let the eqn of normal to the parabola y2=4ax be y=mx-2am-am3 since it is common to the other parabola x2=4a(mx-2am-am3) x2-4amx+8a2m+4a2m3=0 since this normal cuts the parabolas at three distinct points m will have three distinct values thus D=0 16a2m2-4(8a2m+4a2m3)=0 16a2m2-32a2m-16a2m3=0 16a2m(m-2-m2)=0 thus u get m=0 , m2-m+2=0 ............{here as u can see discriminant for this eqn is negetive} But if u take the eqn as m^2-m-2=0. then u get m=-1,2 thus u will get ur correct answer that is why iam getting a bit confused man
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28 Jan 2008 19:14:04 IST
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hey raulrag,how come the normal cuts the parabola in 3 distinct pts, that is not possible..
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28 Jan 2008 19:20:15 IST
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hey sorry
that reason is not appropiate but steps are correct
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