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.(adsbygoogle = window.adsbygoogle || []).push({});

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]

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# Why Repeated Roots instead of Two Distinct Roots

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