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Karthik M

PNP' is the double ordinate of the parabola. Prove that the locus of intersection of the normal at P and the straight line through P' parallel to the axis is the equal parabola y²=4a(x-4a).

This is what I did :

let the point P be (at², 2at). Now, let the normal and the line || to the axis intersect at some Q.
y co ordinate of P' = -2at (As it is double ordinate) So equation of the line is y=-2at.
t = -y/2a.
Equation of normal at P is y+xt = at³+2at.

Putting t = -y/2a, we get: (as the normal meets this line)

y-xy/2a = -ay³/8a³ - y

1-x/2a = -ay²/8a³-1 On further reducing this, I don't get the answer.

But here in the solution, my sir has taken the equation of the normal as
y+xt=at²+at³ . How is that the equation of the normal? Is there anything wrong in my method?

sankydreams

Dude ur sir and u both seem to be wrong
U r doing the ques in hurry
U r right till eqn of normal, i m beginning after that

let pt of intersection of normal and line be (h,k)

putting y = -2at in eqn of normal (they intersect)
-2at = at^3 + 2at - xt
on solving this eqn u get
x = 4a + at^2 = h
k ofcourse equals to -2at
k = -2at
=> t = -k/2a

Putting this value of t in h = 4a + at^2
u get
h = 4a + (k^2)/4a

solve this and get the answer
Rate if this helps u...

Karthik M

Yea... I got the answer that day itself. I realized my folly. Anyway, I' have rated u for your efforts.

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