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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Parabola
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kane (2188)

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prove that the chord of y2=4ax whose equation is y-root2 x+4a root2=0 is a normal to it and its length is 6 root3 a.

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rhd92781 (686)

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hope u know these eqns:
normal to y^2=4ax at point t(at^2, 2at) is
y+tx=2at+at^3
comparing it with y-rt(2)x=-4rt(2)a we get t=-rt(2) and this value of t satisfies 2at+at^3=-4art(2)
hence it is a normal to the parabola (1)
the normal will meet the parabola again at t2=-t-2/t (u shud remember this)
t2 = 2rt(2)
so two points where the normal meets the parabola r (2a, -2rt(2)a) and (8a, 4rt(2)a) (since the points r at^2, 2at)
distance btw them is 6art(3)

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kane (2188)

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comparing it with y-rt(2)x=-4rt(2)a we get t=-rt(2) and this value of t satisfies 2at+at^3=-4art(2)
just explain this line

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rhd92781 (686)

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y+tx=2at+at^3
y-rt(2)x=-4rt(2)
on comparing coeffs of x
t=-rt(2)
so 2at+at^3 = -2rt(2)a-2rt(2)a = -4rt(2)a (we put t=-rt(2) here)
this means the two eqns r identical for a value of t

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But still I can do something;
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I will not refuse to do the something that I can do.
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budku007 (396)

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The equation in slope form of a normal to y^2=4ax is y=mx-2am-am^3 (learn this). Comparing it with y=mx+c we get condition for normality as c=-2am-am^3.For the given equation c=-4a root2 and m=root2.Putting given m in condition of normality we get -4root2a which is equal to c. therefore this equation is a normal. Now using the formula of the length of chord = 4/m^2{(1+m^2)*(a*(a-mc))}^1/2. And substituting a, m and c we get the required result.
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budku007 (396)

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got the answer using this??

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Anyone who has never made a mistake has never tried anything new
You can never solve a problem on the level on which it was created.
When you are courting a nice girl an hour seems like a second. When you sit on a red-hot cinder a second seems like an hour. That's relativity

Purpose of solving a problem is not simply to get the answer(the answer is only an evidence) but to develop your thinking ability
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