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Eqt of the parabola question

  1. Mar 7, 2004 #1
    I've just finished my pure h/work but the final question has me a bit baffled because in lectures we've only been dealing with parabols where the focus is like (a,0) and the directrix is just y=k where k is a real number. However, for this question we've to find the eqt of the parabola whose focus is at the point (1,1) and whose directrix has the eqt x + y = 0

    I've drawn it out so that I've chosen an arbitrary point (x', y') and then worked out the distance between this point and the focus to be the square root of (x' - 1)^2 + (y' - 1)^2

    For working out the distance between the point and the directrix I've got the modulus of (x' + y')/sqrt2

    I've then set these two equal to each other by the definition of the parabola and got x^2 + y^2 -2xy -4x- 4y +4=0 as the eqt of the parabola. Can someone confirm if hi is is correct or not?

    The second part of the question is that if the line ax + by+ c =0 is a tangent to the parabola, then show that the line bx + ay +c=0 is also one of its tangents. I'm really not sure how to go about this one. I thought maybe you could differentiate the eqt of the parabola and then diferentiate the eqts of the tangents and work from there, but it's going pear-sharped. Can anyone point me in the right direction? Thanks!
     
  2. jcsd
  3. Mar 7, 2004 #2

    matt grime

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    Yes, the equation looks right

    as for the second.

    find the tangent to the curve at some point in a general form. let the equation be ax+by+c=0. the gradient of this line is what, in terms of a and b? now find if there is a point on the curve where the tangent has gradient the reciprocal of what yo'uve found. what does that tell you about the equation of the tangent here? How much like the first equation is it?
     
  4. Mar 7, 2004 #3
    I got he gradient of ax+by+c=0 to be -a/b . What do you mean by the reciprocal?
     
  5. Mar 7, 2004 #4

    matt grime

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    if you want a slicker way of doing it:

    the equation is totally symmetric in x and y. if (u,v) is on the curve, so is the point (v,u)

    What is the tangent vector at (u,v) (and at (v,u))

    hence what are the (vector) equations of the tangent lines
     
  6. Mar 7, 2004 #5

    matt grime

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    I mean, what is the gradient of the line bx+ay+c=0?

    it's -b/a

    which is the reciprocal of -a/b

    (reciprocal of x is 1/x)

    find the two points on the curve whose tangents have these slopes. what are the intercepts?
     
  7. Mar 7, 2004 #6
    Hang on the reciprocal of -a/b would be -b/a, right? And that's the graident of bx + ay+c=0 So am I supposed to set these gradients equal the eqt of the parabola differentiated with respect to x? It's just easier for me thais way because I've a fear of vectors being used like that after having a v.mad Russian lecturer last semester who was barmy about vectors! ;)
     
  8. Mar 7, 2004 #7

    matt grime

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    Let (u,v) be a point on the parabola. Find the gradient at (u,v) let this be r

    then the equation is y=rx+c of the tangent for some c.

    you want to show that there is a point on the curve whose tangent has equation

    y=x/r+d

    first find where on the curve the slope is 1/r

    now find the intercepts c and d. nor rearrange

    ry=x+rd,

    so you need rd=c
     
    Last edited: Mar 7, 2004
  9. Mar 7, 2004 #8
    When you say intercepts you mean the y axis intercepts?
     
  10. Mar 7, 2004 #9
    y=x/r+d

    first find where on the curve the slope is 1/r

    How do I do that? Would I set 1/r equal to the gradeint of the parabola?
     
  11. Mar 7, 2004 #10

    matt grime

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    yes, intercept is y axis intercept, and finding where the slope is 1/r means setting the gradient of the parabola to be 1/r.
     
  12. Mar 7, 2004 #11
    So in this instance I'm actually setting it to be -b/a and seeing what I get out of it?
     
  13. Mar 7, 2004 #12

    matt grime

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    That would depend on where you are starting from.

    1. If you are presuming that ax+by+c=0 is a tangent line at the point (u,v) you must demonstrate there is a point where the gradient is -b/a, for then there the tangent line has equation

    y=-bx/a+d,

    rearranging we see

    bx+ay-bd=0,

    you must now find -bd and show it is c.

    How you do that is up to you.




    2. Start from scratch. Throw away what you've written so far.

    Let (u,v) be a point on the parabola.

    Find the equation of the tangent line at (u,v) in terms of u and v.

    write it in the form y=rx+c

    now find where the parabola has slope 1/r, there the tangent line has equation

    y=x/r+d

    rearrange and prove what the question asked you to do.


    Pick one of these methods and stick to it. Tell me which you are doing, where yo have got to with it.
     
  14. Mar 7, 2004 #13

    matt grime

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    Ok, here's my solution


    we see that the slope of the parabola is given by

    [tex]\frac{dy}{dx}=\frac{2+y-x}{y-x-2}[/tex]


    from which we see that the gradient at a point (u,v) is the reciprocal of the gradient at at the point (v,u) which also lies on the parabola by the symmetry of x and y in the equation defining the parabola

    assume this is not zero or infinity

    So the tangents are y=rx+c and y=x/r+d resp, for some c and d.

    we know that the lines go through the points (u,v) and (v,u) resp so we see

    v=ru+c and u=v/r+d so that c = rd

    hence

    y=rx+c is a tangent line iff ry=x+c is a tangent line, as required (we may divide through the original equation as in the question by a, assuming a is not zero).

    The case a=0 is left to you.
     
  15. Mar 7, 2004 #14
    1. If you are presuming that ax+by+c=0 is a tangent line at the point (u,v) you must demonstrate there is a point where the gradient is -b/a, for then there the tangent line has equation

    y=-bx/a+d,

    rearranging we see

    bx+ay-bd=0,

    you must now find -bd and show it is c.

    How you do that is up to you.

    Okay, this is the one I guess I went with. I set -a/b equal to the eqt of the parabola and in the end I got the following-

    y(b+a)= x(b+a) +2b - 2a

    so the y-intercept is 2b -2a


    Then I set -b/a equal to the eqt of the parabola and got the following-

    y(a+b)= x(a+b) -2a + 2b

    so the y-intercept is -2a + 2b


    Since we're told that the eqt of the tangent is ax + by + c=0 and the one we're trying to show is bx +ay +c =0, we can rearrange these both so that we have-

    c = -ax -by (for the eqt we are told)

    c = -bx -ay (for the eqt we are trying to show)

    From what was previously worked out-

    c= 2b-2a for the eqt we know and

    c= -2a +2b for the eqt we're trying to show

    So from these you could see that x=2 and y=-2 for both.

    Now I've no idea why I even went down that path. Think I'll just concentrate on your solution!
     
  16. Mar 7, 2004 #15
    You see when you say the gradient of the point (u,v) is the reciprocal is the gradient at the point v,u how does that work?

    The grad at u,v is (2 + v -u) / (v - u -2)

    And the grad at v,u is (2 + u - v )/ (u-v -2)

    So its reciprocal would be-

    (u - v - 2) / (2 + u - v)

    Spo I'm not sure where I've gone wrong here. Sorry.[b(]
     
  17. Mar 7, 2004 #16

    matt grime

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    multiply top and bottom in the last line by -1, and you get what you want.
     
  18. Mar 7, 2004 #17
    Thanks, I'm so stupid for not seeing that. I fear my brain has shut down after too much Maths this weekend.
     
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