find the point(s) on the parabola x=y^2 closest to the point (0,3)

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find the point(s) on the parabola x=y^2 closest to the point (0,3)

Mathematics
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the closest point to any spot is perpendicular to it......right?
lets take the top part of this; y = sqrt(x) does fine...
the slope of the line of the derivative and the perpendicular line passsing thru (0,3) need to be minimized.....

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y = [1/(2sqrt(x))]x +b 0 = -6sqrt(3) + b b = 6sqrt(3) x y = --------- + 6sqrt(3) 2sqrt(x) .....................and................ y = sqrt(x) x sqrt(x) = --------- + 6sqrt(3) 2sqrt(x) x sqrt(x) - --------- = 6sqrt(3) 2sqrt(x) x 2(sqrt(x))sqrt(x) - --------- = 6sqrt(3) 2sqrt(x) 2x - x --------- = 6sqrt(3) 2sqrt(x)
I sure hope this is right lol x = 12sqrt(3x) x/12 = sqrt(3x) (x/12)^2 = 3x (1/144)x^2 -3x = 0 x((1/144)x - 3) = 0 x = 0 or x= 3/144 = 1/46
we know the distance between 0 and 3 =3, so lets do some tests with numbers close to 1/46. (1/46, sqrt(1/46)) (0,3) sqrt[ (0-1/46)^2 + (3-sqrt(1/46))^2) ] = 2.85264
when x = 1/45.9 we get: (0,3) (1/45.9,sqrt(1/45.9)) sqrt[ (1/45.9 -0)^2 + (sqrt(1/45.9) -3)^2 ] = 2.85248072 x.xxx64088 < x.xxx48072
at x = 1/46.1 we get this: 2.85280052 x.xxx80052 x.xxx64088
great, now I gotta figure out what I did wrong lol....thanx :)
1/2 > 1/3 so the bigger the number on the bottom, the smaller it gets.....
.500 > .333 so the smaller the decimal is, the smaller the number.....
x.xxx80052 > x.xxx64088 x.xxx64088 > x.xxx48072 so we need to work to smaller x values.... maybe
hah!!.. writing it on paper is so much easier lol... x = 9 or x = 1 at x = 9 y = 3 ....................... (0,3) (9,3) sqrt( [9-0]^2 + [3-3]^2) = 9 ........................................... (0,3) (1,1) sqrt([1-0]^2+[1-3]^2) sqrt(1 + 4) = sqrt(5) sqrt(5) = 2.2360679774997896964091736687313 the poits should be (1,1) and (1,-1)

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