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Thread: shortest chord

  1. #1
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    shortest chord

    What normal to the curve y = x^2 forms the shortest chord?

    This is what I did:
    I used parametric form of parabola and did stuff, put it in distance formula.. d/dt = 0 it, I'm getting a complicated polynomial ( 7th degree ) after doing that..
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  2. #2
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    Quote Originally Posted by ice_syncer View Post
    What normal to the curve y = x^2 forms the shortest chord?

    This is what I did:
    I used parametric form of parabola and did stuff, put it in distance formula.. d/dt = 0 it, I'm getting a complicated polynomial ( 7th degree ) after doing that..
    I don't believe that this problem has a neat solution. I tried it for the parabola $\displaystyle y^2=4ax$, where the general point has parametric form $\displaystyle (at^2,2at)$. The normal at this point has equation $\displaystyle y-2at = -t(x-at^2)$, and it meets the parabola at the point with parameter $\displaystyle s = -t-\frac2t$. So if d is the length of the chord then $\displaystyle d^2 = (at^2-as^2)^2 + (2at-2as)^2$, which works out as

    $\displaystyle d^2 = a^2\Bigl(4t^2 + 24 + \dfrac{36}{t^2} + \dfrac{16}{t^4}\Bigr)$.

    Differentiate that to find that the turning point (which must be a minimum) occurs when $\displaystyle t^6 - 9t^2-8 = 0$. You can factorise that as $\displaystyle t^6 - 9t^2-8 = (t^2+1)(t^4-t^2-8)$. The second factor is a quadratic in $\displaystyle t^2$, and the only positive root is $\displaystyle t^2 = \frac12\bigl(1+\sqrt{33})$. Therefore the minimum occurs when $\displaystyle t = \sqrt{\frac12\bigl(1+\sqrt{33})} \approx 1.836377...$. If you now substitute the value for $\displaystyle t^2$ into the formula for $\displaystyle d^2$, taking a=1/4, and finally take the square root to get d, then you will have the length of the shortest chord.
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  3. #3
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    I was hoping for some answer that uses geometry and trigonometry to find the length, rather than using distance formula
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