shortest chord

**ice_syncer** · November 10th 2010, 10:17 PM

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..

**Opalg** · November 11th 2010, 04:49 AM

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.

**ice_syncer** · November 15th 2010, 01:07 AM

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

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