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Tangent ad normal parabola

  1. Nov 29, 2013 #1

    utkarshakash

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    Tangent and normal parabola

    1. The problem statement, all variables and given/known data
    P is a point 't' on the parabola y^2=4ax and PQ is a focal chord. PT is a tangent at P and QN is a normal at Q. The minimum distance between PT and QN is equal to

    2. Relevant equations

    3. The attempt at a solution

    I think minimum distance will occur when the chord PQ is parallel to Y-axis. In other words, the normal at Q will be parallel to the tangent at P. In that case I can apply the formula for calculating the distance between two parallel lines. But this does not give correct answer. I also thought about 0 but I don't think these questions have such simple answers.
     
    Last edited: Nov 29, 2013
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  3. Nov 29, 2013 #2

    Mentallic

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    Unless you have some reason to believe that, I'd consider it to be a big stab in the dark.

    My approach was rather tedious, but I always remembered these parametric parabolas to be like that, so there might not be a much simpler approach.
    I began with finding the equations for PT and QN, then deriving the distance formula for the distance between two parallel lines, I had my distance in terms of y0 (which is the y-value of the point P) and the constant a. Then I took the derivative of the distance d with respect to y0 and equated that to 0.

    The answer isn't something that you can simply guess.
     
  4. Nov 29, 2013 #3

    utkarshakash

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    But how can you say that the tangent at P and normal at Q will be parallel to each other?
     
  5. Nov 29, 2013 #4

    Mentallic

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    Well I proved it for myself, but didn't bother mentioning it because I assumed you accepted it to be true for whatever reason since you didn't seem to question it in your first post.

    If you find PT and QN then you simply show that their gradients are equal.
     
  6. Nov 29, 2013 #5

    utkarshakash

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    Never mind. I figured it out. But my real concern lies in finding the 't'. I followed your approach and reached a point where I get a 7 degree equation in t. Do I really need to solve for t? I assume not because the answer has a as well as 't' in it. If there were a particular 't' for minimum distance, then the answer would be free from 't', but that's not the case here.
     
  7. Nov 29, 2013 #6

    Mentallic

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    So you have an expression that's

    [tex]d=f(t,a)[/tex]

    where d is the distance, t is the parameter, and a is the focus, and d is a 7th degree polynomial in terms of t?

    Given two parallel lines

    [tex]y_1=mx+b[/tex]

    [tex]y_2=mx+c[/tex]

    What is the distance between them?
     
  8. Nov 29, 2013 #7

    utkarshakash

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    [itex]\dfrac{b-c}{\sqrt{1+m^2}}[/itex]

    In this case the distance comes out to be

    [itex]\dfrac{at^2 + \frac{2a}{t} + \frac{a}{t^3}}{\sqrt{1+\frac{1}{t^2}}}[/itex]

    Differentiating and setting it to zero gives me

    [itex]2t^4+3t^2-2t- \frac{3}{t} - \frac{2}{t^3}=0[/itex]
     
  9. Nov 29, 2013 #8

    Mentallic

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    Ok so I see you've used the parameters [itex]P(at^2,2at)[/itex]. This does ring a bell from school.

    Technically it would be the absolute value of that, but we can easily fix b>c and it doesn't really change anything.

    I get at rather than at2 for that first term in the numerator. Do you agree that PT is

    [tex]y-2at=\frac{1}{t}(x-at^2)[/tex]

    and QN is

    [tex]y+\frac{2a}{t}=\frac{1}{t}(x-\frac{a}{t^2})[/tex]
     
  10. Nov 30, 2013 #9

    utkarshakash

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    Yes. It was a silly mistake from my side.
    So the final equation changes to

    [itex]-(1+t^2)\left( 1 - \dfrac{2}{t^2} - \dfrac{3}{t^4} \right) = \left( 1 + \dfrac{2}{t^2} + \dfrac{1}{t^4} \right)[/itex]

    I hope I'm right this time.
     
  11. Nov 30, 2013 #10

    Mentallic

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    What is that? The derivative of d with respect to t? There are no real solutions for t when setting that to 0, which wouldn't make any sense in this context since it explains where the minimum or maximum of the distance must occur.

    [tex]\frac{\partial}{\partial t}\left( \frac{at+\frac{2a}{t}+\frac{a}{t^3}}{\sqrt{1+\frac{1}{t^2}}}\right) = \frac{a\sqrt{1+\frac{1}{t^2}}\left( t^2-2\right) }{t^2}[/tex]
     
  12. Nov 30, 2013 #11

    utkarshakash

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    Can you explain what have you done here?
     
  13. Nov 30, 2013 #12

    Mentallic

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    I used the product rule.

    [tex]\frac{\partial}{\partial t}\left( \frac{at+\frac{2a}{t}+\frac{a}{t^3}}{\sqrt{1+\frac{1}{t^2}}}\right)= a\cdot \frac{d}{dt}\left( \frac{t+\frac{2}{t}+\frac{1}{t^3}}{\sqrt{1+\frac{1}{t^2}}}\right)[/tex]

    Let

    [tex]u=at+2at^{-1}+at^{-3} [/tex]

    [tex]v= (1+t^{-2})^{-1/2}[/tex]

    Then

    [tex]u'v+uv' = (a-2at^{-2}-3at^{-4})(1+t^{-2})^{-1/2}+(at+2at^{-1}+at^{-3})\cdot -1/2\cdot (1+t^{-2})^{-3/2}\cdot -2t^{-3}[/tex]

    [tex]=\frac{a-2at^{-2}-3at^{-4}}{(1+t^{-2})^{1/2}}+\frac{at+2at^{-1}+at^{-3}}{t^3(1+t^{-2})^{3/2}}[/tex]

    Now we'll factor out anything that's in common and anything that will make things more simple for us. We'll also factor out a t-2 from the denominators so we're only dealing with polynomials.

    [tex]=\frac{a\sqrt{1+t^{-2}}}{t^2}\left(\frac{t^2-2-3t^{-2}}{t^{-2}(t^2+1)}+\frac{t^3+2t+t^{-1}}{t^3(t^{-2})^2(t^2+1)^2}\right)[/tex]

    [tex]=\frac{a\sqrt{1+\frac{1}{t^2}}}{t^2}\left(\frac{t^4-2t^2-3}{t^2+1}+\frac{t^4+2t^2+1}{(t^2+1)^2}\right)[/tex]

    I'm sure you can take it from here. The numerators factor nicely.
     
  14. Nov 30, 2013 #13

    utkarshakash

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    I'd really like to thank you for the effort you've made in posting all the steps of differentiation. Since you already told that setting the derivative to 0 wouldn't give me a real 't' then how can this derivative help me?
     
  15. Dec 1, 2013 #14

    Mentallic

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    Sorry, I didn't mean to insult your intelligence, but you did ask me to explain how I got the result that I did. If you simplify the part in the brackets that I left for you to do, you'll find it's equal to [itex]t^2-2[/itex] which means that when you set the result equal to 0, [itex]t=\pm \sqrt{2}[/itex].

    There is no real t that lets this expression equal 0. I'm also still not sure what this was supposed to be.
     
  16. Dec 1, 2013 #15

    utkarshakash

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    Don't take me wrong. I was only appreciating you :smile:Finally, I arrived at the correct answer. Thanks for helping me out with this difficult question.
     
  17. Dec 1, 2013 #16

    Mentallic

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    Oh, I guess I was sensing sarcasm where there wasn't any :biggrin:

    You're welcome! I do wonder if there is a simpler solution though.
     
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