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Why Repeated Roots instead of Two Distinct Roots

  1. Jun 10, 2012 #1
    Greetings all. I hope it's OK to post here. My issue here is with the theory and not with the actual algebra or calculus.

    I understand this calculus question on parametric curves except why there must be a double root instead of just a repeated root in the last part. Please see the red question mark below.

    I understand that for the values of [itex] t [/itex] on curve C which two normals can be drawn to, the equation [itex]0=at^3 - th + 2at-k[/itex] will have two roots. But why must these roots be identical and so repeated?

    Thank you all.

    Original question

    [itex]\boxed{\text{Recall that }{{x}^{3}}+px+q=0\text{ has exactly one real solution if }p\ge 0. \\
    \text{A parabola }C\text{ is given parametrically by }x=a{{t}^{2}},\text{ }y=2at\text{ (}a>0). \\
    \text{Find an equation which must be satisfied by }t\text{ at points on }C\text{ at which the normal } \\
    \text{passes through }(h,k).\text{ Hence show that if }h\le 2a\text{, exactly one normal to }C\text{ will pass through }(h,k). \\
    \text{Find, in Cartesian form, the equation of the locus of the points from which exactly two normals can be drawn to }C.} [/itex]

    Original solution

    [itex] y = 2at, x = at^2 \text{ so } \dot{y} = 2a, \dot{x} = 2at \implies \displaystyle -1/\frac{dy}{dx} = -t[/itex]. Therefore the equation of a normal will be [itex]\displaystyle \frac{y - 2at}{x-at^2} = -t \iff y = -t x + at^3+2at[/itex]

    therefore we have [itex]0=at^3 - th + 2at-k[/itex]. By considering the coefficient of t, for there to be one solution [itex]2a -h \ge 0 \Longleftrightarrow 2a \ge h[/itex]

    [itex]0=at^3 - th + 2at-k[/itex] if two normals can be drawn then we have a repeated root. (?)

    Therefore [itex]3at^2 +(2a-h)=0 \implies \displaystyle t = \pm \sqrt{\frac{h-2a}{3a}}[/itex]

    So [itex]\pm \sqrt{\frac{h-2a}{3a}} \left ( a \left ( \frac{h-2a}{3a} \right ) + 2a - h \right ) -k =0 \implies \displaystyle k^2 = \frac{4(h-2a)^3}{27a}[/itex]
  2. jcsd
  3. Jun 10, 2012 #2


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    If a polynomial equation with real coefficients has a complex root, then the complex conjugate of that root is also a root. That means that a cubic equation has either three real roots or two complex and one real root. In particular, it is impossible for a cubic equation to have two real roots and one complex root.

    If a cubic equation has two distinct roots, then it must have three real roots so if there are only two distinct roots, the third root must be the same as one of the first twol.
  4. Jun 10, 2012 #3
    Thank you HallsofIvy. I'm going sum up your answer. Please let me know if it's right or wrong. Did I miss anything?

    If [itex] t [/itex] gives a point on the locus (NOT C) which two normals to C can be drawn to, there CAN in fact be two different values of [itex] t [/itex] that satisfy [itex] 0=at^3 - th + 2at-k[/itex]. So this equation can have two distinct real roots.

    But as you explained, because a cubic cannot have only two real roots, therefore there must be 3 real roots to [itex] 0=at^3 - th + 2at-k[/itex]. From the paragraph above, we had only 2 distinct real roots. Therefore, the 3rd one must be a repeated root.

    So in total, we have 2 distinct real roots here and 1 repeated root, for a total of 3. We don't know which of the 2 distinct real roots is repeated?
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