# Thread:

1. ## 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..

2. Originally Posted by ice_syncer
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 $y^2=4ax$, where the general point has parametric form $(at^2,2at)$. The normal at this point has equation $y-2at = -t(x-at^2)$, and it meets the parabola at the point with parameter $s = -t-\frac2t$. So if d is the length of the chord then $d^2 = (at^2-as^2)^2 + (2at-2as)^2$, which works out as

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

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

3. I was hoping for some answer that uses geometry and trigonometry to find the length, rather than using distance formula
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