Find Minimum Value of T^2 in Normal to Parabola y^2 = 4ax at (at^2, 2at)

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The discussion centers on finding the minimum value of T^2 for the normal to the parabola y^2 = 4ax at the point (at^2, 2at), which intersects the parabola again at (aT^2, 2aT). The equation of the normal is derived and simplified to form a quadratic equation, t^2 + tT + 2 = 0. The minimum value of T^2 is determined by analyzing the quadratic's properties, particularly its discriminant, to ensure real roots exist. The conclusion reached is that the minimum value of T^2 is 8, which arises from the condition that the quadratic must have real solutions. This approach effectively connects the minimum value of T^2 with the geometric properties of the parabola and its normal.
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Homework Statement


If the normal to the parabola y^2 = 4ax at the point (at^2 , 2at) cuts the parabola again at (aT^2, 2aT), then minimum value of T^2 is
ans: 8

I got the answer but I don't know why it should be the answero_O?

Homework Equations


Equation of normal to the parabola in parametric form can be written as
y-2at = -t(x-at^2)

The Attempt at a Solution


So, in the above equation, I substituted (aT^2, 2aT) as this point will lie on the normal.
2aT - 2at = -t(aT^2 - at^2)
On simplifying,
t^2 + tT + 2 = 0

From here on, I don't quite understand why what I did works.
Minimum value of this quadratic will be at (-b/2a, f(-b/2a) ).
i.e., (-T/2, f(-T/2) )
f(-T/2)= (- T^2 / 4 ) + 2
I equate ( -T^2 / 4 ) + 2 to 0.
I get T^2 = 8.
What did I just do?
 
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erisedk said:

Homework Statement


If the normal to the parabola y^2 = 4ax at the point (at^2 , 2at) cuts the parabola again at (aT^2, 2aT), then minimum value of T^2 is
ans: 8

I got the answer but I don't know why it should be the answero_O?

Homework Equations


Equation of normal to the parabola in parametric form can be written as
y-2at = -t(x-at^2)

The Attempt at a Solution


So, in the above equation, I substituted (aT^2, 2aT) as this point will lie on the normal.
2aT - 2at = -t(aT^2 - at^2)
You can write this as 2a(T- t)= -at(T- t)(T+ t)
T- t= 0 only when T= t so NOT another point of intersection. Since T- t is not 0, we can divide by a(T- t) to get
2= -t(T+ t) or t^2+ Tt+ 2= 0.

On simplifying,
t^2 + tT + 2 = 0

From here on, I don't quite understand why what I did works.
Minimum value of this quadratic will be at (-b/2a, f(-b/2a) ).
i.e., (-T/2, f(-T/2) )
Complete the square: t^2+ Tt + 2= t^2+ Tt+ T^2/4- T^2/4+ 2= (t+ T/2)^2- (T^2/4-2).. Since a square is never negative, this will be minimum when t+ T/2= 0 or t= -T/2. In that case, the equation becomes -(T^2/4- 2)= 0 so that T^2/4= 2, T^2= 8.

f(-T/2)= (- T^2 / 4 ) + 2
I equate ( -T^2 / 4 ) + 2 to 0.
I get T^2 = 8.
What did I just do?
 
HallsofIvy said:
(t+ T/2)^2- (T^2/4-2).. Since a square is never negative, this will be minimum when t+ T/2= 0 or t= -T/2.
t^2+ Tt + 2 is just a relation between two points. Why does minimising it work? For ex: If I had the function f(x)=x^2 + 3x + 10 , and I was asked to find its minimum value, I would do what you did above, figure out the minima. This gives me the least value of a certain CURVE, in this case, of a parabola. How does finding the minimum value of t^2+ Tt + 2 give me the minimum value of T^2 ? Just tell me WHY it works, I get how you did it, which is also basically what I did.
 
erisedk said:
t^2+ Tt + 2 is just a relation between two points. Why does minimising it work?
t^2+ Tt + 2 =0 is the conclusion you have come to in order to satisfy the information you were given (i.e. the point is on the normal).
The question asked for the minimum value of T. You could read that as the minimum value of T for which the quadratic has real roots.
i.e. ##\sqrt{T^2 -8} \geq 0##.
 
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