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Tangent drawn to parabola

  1. Aug 1, 2015 #1
    I know a tangent drawn on parabola having equation like
    y^2=4ax is
    y=mx+(a/m)
    which provides c=a/m
    Then how is it going to turn for equation like x^2=4ay?

    From my derivation it will be like -c=am^2
    when the equation of tangent is y=mx+c.

    The derivation comes from the following:
    y=mx+c
    or, x=(y/m)-(c/m)

    So, comparing with the tangent on
    y^2=4ax we get

    -(c/m)=a/(1/m)
    that is -c= am^2

    But the problem arises when in a question saying find the common tangent on y^2=4ax and x^2=4ay, the solution was made taking the tangents for each parabola as
    y=mx+a/m and
    x= my + a/m respectively.

    Shouldn't the later one be x=(y/m)-(c/m)
    i.e x=(y/m)-am ?
     
  2. jcsd
  3. Aug 1, 2015 #2

    HallsofIvy

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    What you say,
    "I know a tangent drawn on parabola having equation like
    y^2=4ax is
    y=mx+(a/m)
    which provides c=a/m"

    Simply doesn't mean anything because you have not said what "m" means!

    Please rewrite this, telling us what "m" is, so that it makes sense.
     
  4. Aug 1, 2015 #3

    Mentallic

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    Homework Helper

    If choosing various m gave all the tangents to the parabola, then it'd make sense. This is not the case however.
     
  5. Aug 2, 2015 #4
    It's a general straight line equation, where m is the slope of a tangent
     
  6. Aug 2, 2015 #5
    Yes, it is the case.

    If the tangent is y=mx+c

    where m=slope of the line
    c=intercept it cuts on y axis

    And it is a tangent to a general parabola of equation

    y^2=4ax

    then c=a/m
     
  7. Aug 3, 2015 #6

    Mentallic

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    Yes, sorry, you're right. I made a quick base-case graph check and it went wrong somewhere along the way.

    The derivation for your new tangent problem is very simple to derive. Since you've already correctly found that
    [itex]y=mx+a/m[/itex] is tangent to [itex]y^2=4ax[/itex]
    then symmetrically,
    [itex]x=my+a/m[/itex] is tangent to [itex]x^2=4ay[/itex]


    You'll need to justify this last step. I can't follow that line of thought.
     
    Last edited: Aug 3, 2015
  8. Aug 3, 2015 #7
    In this line you are taking x=my+c as a tangent of the x^2=4ay

    Here, m is not the slope, c is not the intercept of y-axis.

    Here m= 1/(slope) and
    c=-(intercept/slope)

    I just put in case of m, 1/m and in place of c, -(c/m) according to the conventional meaning of m and c.

    I just tried it on paper. Both are actually same, denoting the same meaning until you are confused what m and c means in which equation.
     
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