A Normal and a Parabola (Posted on 2009-12-02)

	Submitted by Brian Smith
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k=1/2, area=4/3 The slope of the tangent at (k,k^2) is 2k, so the slope of the normal is -1/(2k). Then the equation of the normal is y = -x/(2k) + k^2 + 1/2. The second point the normal intersects y=x^2 is ( -k + -1/(2k), k^2 + 1 + 1/(4k^2) ). The area of the closed region is equal to the integral of the difference of the equations of the normal and parabola from x=-k-1/(2k) to k. =Integ{-k-1/(2k), k} [(-x/(2k) + k^2 + 1/2) - x^2] dx The easiest way to evaluate this integral is to use Simpson's rule which is exact for parabolas. To do that one more point is needed at x=[(k)+(-k-1/(2k))]/2 = -1/(4k) The three points to be used are: (k, 0) (-1/(4k), k^2 + 1/2 + 1/(16k^2) (-k-1/(2k), 0) Then the integral evaluates to Area = ( 2k+1/(2k) )/6 * [1*0 + 4*(k^2 + 1/2 + 1/(16k^2)) + 1*0] This simplifies to Area = 1/6 * [2k+1/(2k)]^3 2k+1/(2k) and its cube are at a minimum when k=1/2 (remember k>0). Then the area is 1/6 * 2^3 = 4/3

	Subject	Author	Date
	The required value of k is...	K Sengupta	2023-08-01 23:05:49
	re(2): solution error found	Daniel	2009-12-02 22:08:16
	re: solution error found	Ady TZIDON	2009-12-02 14:43:34
	re: solution	Charlie	2009-12-02 12:43:05
	solution	Daniel	2009-12-02 12:13:35