Question

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

Answer 1

the closest point to any spot is perpendicular to it......right?

Answer 2

lets take the top part of this; y = sqrt(x) does fine...

Answer 3

the slope of the line of the derivative and the perpendicular line passsing thru (0,3) need to be minimized.....

Answer 4

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)

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

at x = 1/46.1 we get this: 2.85280052


x.xxx80052
x.xxx64088

Answer 9

great, now I gotta figure out what I did wrong lol....thanx :)

Answer 10

1/2 > 1/3 so the bigger the number on the bottom, the smaller it gets.....

Answer 11

.500 > .333 so the smaller the decimal is, the smaller the number.....

Answer 12

x.xxx80052 > x.xxx64088

x.xxx64088 > x.xxx48072 so we need to work to smaller x values.... maybe

Answer 13

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)