Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

It seems that something is wrong either with this proof or my understanding.

  1. Oct 23, 2011 #1
    Here i am asking all the things about parabola. I think something is wrong with proof where we prove condition of tangency.
    It is done shown(in my book)
    [itex]y^2=4ax[/itex] //equation of general parabola
    let us assume that line [itex]y=mx+c[/itex] is passing through parabola.
    points where it will cut parabola.
    solving this equation.
    line will touch parabola at one point if of the roots of this equation are equal.
    It means it’s discriminant is zero.
    => [itex]16a^2=16amc[/itex]
    [itex]a=0 [/itex]; not possible because in that case it will not remain a parabola it became a line y=0.
    this is the require condition for the line to cut the parabola at one point.
    So lets take an example.
    [itex]y=2[/itex]. this line cuts the parabola [itex]y^2-4x=0[/itex] at one point. as we can see on the graphs
    but does it obey the equation proved previously.
    a=0/2. NO. it is not obeying that equation.
    the condition is for line to cut the parabola at one point.y=2 is also a line that cuts parabola at one point but not obeying the condition.
    Last edited: Oct 23, 2011
  2. jcsd
  3. Oct 23, 2011 #2
    To conclude that this equation is a quadratic you need to know that m≠0. If m=0, then this is just a linear equation and will have exactly one solution regardless of what the discriminant is. Since your given line does in fact have m=0, the discriminant need not be zero, and so a is not necessarily equal to mc.
  4. Oct 24, 2011 #3
    problem solved.
    million thanks for answering.
  5. Oct 24, 2011 #4


    User Avatar
    Science Advisor

    By the way, when m is not 0, "touching the parabola at one point" means "tangent to the parabola" which is what the discriminant being 0 gives.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook