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

[leprofece]

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

 

[MarkFL]

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.

 

[leprofece]

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.

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)

 

[MarkFL]

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 $N$?

 

[leprofece]

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?

 

[MarkFL]

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

$$-\frac{dx}{dy}$$

 

[leprofece]

the derivative is
m = 2p/(y+u)²

and normal N = - (y+u)²/2p

 

[MarkFL]

You have:

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

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

 

[leprofece]

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

 

[MarkFL]

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

 

[Fermat1]

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?

 

[MarkFL]

I get a different value for $u$...how did you obtain the value you have?

 

[Fermat1]

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?

 

[MarkFL]

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.

 

[leprofece]

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

 

[MarkFL]

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.

 

[leprofece]

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

 

[leprofece]

How can I get to f(y)=4((p/y)²+1)(y²+1)
my friend?
Could you please tell me??

 

[MarkFL]

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...

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

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

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

 

[leprofece]

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

 

[MarkFL]

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)

 

[leprofece]

I have just derived the expression that we have got before

 

[MarkFL]

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.