What is the intercept of N when MN is minimal in a parabola y2 = 2px?

  • Context: MHB 
  • Thread starter Thread starter leprofece
  • Start date Start date
  • Tags Tags
    Max Parabola
Click For Summary
SUMMARY

The discussion focuses on calculating the intercept of point N when the distance MN is minimized on the parabola defined by the equation y² = 2px. Participants derive the coordinates for points N and M as N( y²/2p, y) and M( u²/2p, u), respectively. They explore the slope of the normal line at point N and utilize calculus to express u as a function of y and p. The final objective function for minimizing the distance MN is f(y) = 4((p/y)² + 1)(y² + 1).

PREREQUISITES
  • Understanding of parabolic equations, specifically y² = 2px.
  • Knowledge of calculus, particularly differentiation and optimization techniques.
  • Familiarity with coordinate geometry and distance formulas.
  • Proficiency in algebraic manipulation and simplification of expressions.
NEXT STEPS
  • Study the properties of parabolas and their normals in coordinate geometry.
  • Learn about optimization techniques in calculus, including critical points and the second derivative test.
  • Explore the use of LaTeX for presenting mathematical expressions clearly.
  • Investigate distance minimization problems in geometry and their applications.
USEFUL FOR

Mathematicians, students studying calculus and geometry, and anyone interested in optimization problems involving parabolas.

leprofece
Messages
239
Reaction score
0
the normal at a point N of the parabole: y2 = 2px to the curve at another point M. calculate the intercept of N when the length of MN is minimal.

Answer sqrt (2) p
 
Physics news on Phys.org
Re: max and min parabole

I would begin be letting point $N$ be $$\left(\frac{y^2}{2p},y\right)$$ and point $M$ be $$\left(\frac{u^2}{2p},u\right)$$. Now, you know what the slope of the line through $M$ and $N$ has to be since this line is normal to the given parabola at $N$, so this will allow you to express $u$ as a function of $y$ and the parameter $p$.

Then you can construct a function representing the distance (or the square of the distance) between $M$ and $N$ which you can minimize. Once you have the particular $y$ critical $y$-value you can then find the root or $x$-intercept of this line.
 
Re: max and min parabole

MarkFL said:
I would begin be letting point $N$ be $$\left(\frac{y^2}{2p},y\right)$$ and point $M$ be $$\left(\frac{u^2}{2p},u\right)$$. Now, you know what the slope of the line through $M$ and $N$ has to be since this line is normal to the given parabola at $N$, so this will allow you to express $u$ as a function of $y$ and the parameter $p$.

Then you can construct a function representing the distance (or the square of the distance) between $M$ and $N$ which you can minimize. Once you have the particular $y$ critical $y$-value you can then find the root or $x$-intercept of this line.

Calculating I got
Slope m = (2p (u-y))/(u2-y2)

Solving para y i must get y = (-2px)/(u + y)

Distance = sqrt ( y2-u2)/2p)2 + (u-y)2)
 
You are correct in that the slope of the line is:

$$m=\frac{2p}{y+u}$$

(I simplified a bit).

Now, can you use the calculus to come up with another expression for the slope using the fact that the line is normal to the parabola at $N$?
 
MarkFL said:
You are correct in that the slope of the line is:

$$m=\frac{2p}{y+u}$$

(I simplified a bit).

Now, can you use the calculus to come up with another expression for the slope using the fact that the line is normal to the parabola at $N$?

It must be -(y+u)/2p ??

How can proceed?
 
No, you need to use differentiation to find the slope of the normal line at $N$. You want to compute:

$$-\frac{dx}{dy}$$
 
the derivative is
m = 2p/(y+u)2

and normal N = - (y+u)2/2p
 
You have:

$$x=\frac{y^2}{2p}$$

So, what is $$-\frac{dx}{dy}$$?
 
MarkFL said:
You have:

$$x=\frac{y^2}{2p}$$

So, what is $$-\frac{dx}{dy}$$?

derivative of x as a function of y
y/p is the derivative
 
  • #10
Correct, but you want the negative of this. And now you have two expressions representing the slope of the normal line and you may equate them and get $u$ as a function of $y$. Then continue following the instructions I gave in my first post. :D
 
  • #11
MarkFL said:
Correct, but you want the negative of this. And now you have two expressions representing the slope of the normal line and you may equate them and get $u$ as a function of $y$. Then continue following the instructions I gave in my first post. :D

I get $u=\frac{-y}{2p^2}-y$ and the distance squared between $N$ and $M$ to be $y^2(2+\frac{1}{2p^2})^2+\frac{y^4(1+\frac{1}{2p^2})^2-p)^2}{2p^2}$ Is this right and how do you proceed?
 
  • #12
I get a different value for $u$...how did you obtain the value you have?
 
  • #13
MarkFL said:
I get a different value for $u$...how did you obtain the value you have?

sorry, I now get $u=(2p^2-y^2)/y)$. How do you minimise the length, which is quite a nasty expression?
 
  • #14
That is closer to the value I have for $u$, but still different. The resulting expression for the square of the distance $\overline{MN}$, while daunting at first, may be algebraically simplified substantially.
 
  • #15
well The distance you have I don't know if it is good but I don't understand where they come from
begining
you wrote y2(1 +1/2p2)2 + or = ?? (y4(1 +1/2p2)2-p2) y2(1 +1/2p2)2
After operating I got y2(1 +1/2p2)2(p2+y2) y2(1 +1/2p2)2/ 2p2

Maybe I am wrong if so, tell me how you got to the u in your last post
 
  • #16
I got:

$$u=-\frac{2p^2+y^2}{y}$$

And so the objective function (the square of $\overline{MN}$) is:

$$f(y)=\left(\frac{y^2}{2p}-\frac{\left(\dfrac{2p^2+y^2}{y}\right)^2}{2p}\right)^2+\left(y+\frac{2p^2+y^2}{y}\right)^2$$

After some algebraic simplification, I obtained:

$$f(y)=4\left(\left(\frac{p}{y}\right)^2+1\right)\left(y^2+1\right)$$

Now, I will leave you to verify this, and then minimize the objective function.
 
  • #17
MarkFL said:
I got:

$$u=-\frac{2p^2+y^2}{y}$$

And so the objective function (the square of $\overline{MN}$) is:

$$f(y)=\left(\frac{y^2}{2p}-\frac{\dfrac{2p^2+y^2}{y}}{2p}\right)^2+\left(y+\frac{2p^2+y^2}{y}\right)^2$$

After some algebraic simplification, I obtained:

$$f(y)=4\left(\left(\frac{p}{y}\right)^2+1\right)\left(y^2+1\right)$$

Now, I will leave you to verify this, and then minimize the objective function.

After an hour trying to get your answer I was not able to get it
I got y3(y+2)/(4p2) + (y4+4p2)(1+16p2)(42y2])- (y +1 +8y2)

So It is a very difficult algebra
 
  • #18
How can I get to f(y)=4((p/y)2+1)(y2+1)
my friend?
Could you please tell me??
 
Last edited:
  • #19
This is what I did:

$$f(y)=\left(\frac{y^2}{2p}-\frac{\left(\dfrac{2p^2+y^2}{y}\right)^2}{2p}\right)^2+\left(y+\frac{2p^2+y^2}{y}\right)^2=$$

$$\frac{1}{4p^2}\left(y^2-\frac{4p^4+4p^2y^2+y^4}{y^2}\right)^2+\left(\frac{2p^2+2y^2}{y}\right)^2=$$

$$\frac{1}{4p^2}\left(\frac{4p^4+4p^2y^2}{y^2}\right)^2+4\left(\frac{p^2+y^2}{y}\right)^2=$$

$$4p^2\left(\frac{p^2+y^2}{y^2}\right)^2+4\left(\frac{p^2+y^2}{y}\right)^2=$$

At this point, I discovered I originally made an error...

$$4\left(p^2+y^2\right)^2\left(\frac{p^2}{y^4}+\frac{1}{y^2}\right)=$$

$$4\left(p^2+y^2\right)^2\left(\frac{y^2+p^2}{y^4}\right)=$$

$$4\left(\left(\frac{p}{y}\right)^2+1\right)\left(p^2+y^2\right)$$
 
  • #20
Thanks now I derived this expression and got
8(1+p2)(y-y2+p2)/(y2)
and I equated to 0
and got y2-y-p2 but I didnt get anything
 
  • #21
leprofece said:
Thanks now I derived this expression and got
8(1+p2)(y-y2+p2)/(y2)
and I equated to 0
and got y2-y-p2 but I didnt get anything

Well, not knowing what you did, I can't really say where you went wrong. :D

I highly recommend for you to take some time to learn how to use $\LaTeX$ to show your work...plain text is very hard to read for all but the simplest of computations and expressions. This will be a great way for you to help us to help you. (Sun)
 
  • #22
I have just derived the expression that we have got before
 
  • #23
leprofece said:
I have just derived the expression that we have got before

Until I see what you did, I can't possibly tell you where you went wrong.
 

Similar threads

High School Calculating shaded area: I'm getting discrepancies between methods
  • · Replies 3 ·
Replies
3
Views
3K
MHB Max and min line equation
  • · Replies 1 ·
Replies
1
Views
1K
Projectile motion with ##N## bounces on the ground
  • · Replies 21 ·
Replies
21
Views
1K
Undergrad Minimize supported weight of a water pipe
  • · Replies 28 ·
Replies
28
Views
3K
MHB Solve Archway/Spire Building | Find Coordinates, Values, & Domain
  • · Replies 1 ·
Replies
1
Views
3K
What is the correct formula for the reduced Chi square?
  • · Replies 3 ·
Replies
3
Views
4K
What is the process for finding the unit normal at every point on a parabola?
  • · Replies 5 ·
Replies
5
Views
9K
MHB What is the limit of the y-intercept as P approaches O on a given parabola?
  • · Replies 1 ·
Replies
1
Views
1K
MHB Max dimensions of box template
  • · Replies 15 ·
Replies
15
Views
4K
Monte Carlo simulation, rejection algorithm
  • · Replies 2 ·
Replies
2
Views
2K