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Normal to parabola

  • Thread starter erisedk
  • Start date
  • #1
374
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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?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
956

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?
 
  • #3
374
7
(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.
 
  • #4
RUber
Homework Helper
1,687
344
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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