Question

HELP PLEASE

Find the minimum distance from the point (4, 2) to the parabola y^2=8x

Answer 1

The square of the distance from the point to the line is
D^2 = (x-4)^2 + (y-2)^2
Let y and x(y) vary and minimize that.
D^2 = [(y^2/8)-4]^2 +(y-2)^2
D^2 = (y^4)/64 -y^2 + 16 +y^2 -4y +4
= y^4/64 - 4y +20
d(D^2)/dy = 0
(y^3)/4 -4 =0
y^3 = 16
y = 2*2^(1/3) = 2.5196
x = 0.79355

Verify (or correct) my work and compute D^2 , and then D

Answer 2

(y^3)/4 -4 =0


isnt that supposed 2 be
(y^3)/16 -4 = 0

Answer 3

Yes. So y^3 = 64 and y = + or - 4

Nice job catching that mistake!

x = 2, either way.

One of the two points may be a relative minimum only

Answer 4

z=2 square root of 2

Answer 5

2sqrt2

Answer 6

eq. 1 - - - > (y^2)=8x - - - > x=(y^2)/8
eq. 2 - - - -> point distance formula
d=[(x2-x1)^2+(y2-y1)^2] ^1/2
input (4,2) and eq. 1 to eq. 2

d=[((y^2)/8 - 4)^2 + (y - 2)^2] ^1/2

differentiate d
let u = ((y^2)/8 - 4)^2 + (y - 2)^2
du = 2((y^2)/8 - 4)^(2-1) ((2y) /8)
+ 2(y - 2)^(2-1) (1)
d= (u) ^1/2
d'=(1/2) (u) ^(1/2 - 1) (du)
d'=du/[2(u)^(1/2)]

then rule of minima maxima equate d' to 0
0 = du/[2(u)^(1/2)]

cross multiplication

0=du

0=2((y^2)/8 - 4)((2y)/8) + 2(y-2)
0=(4y^3)/64 - 2y + 2y - 4
y^3= 64 - - - > y=4

input 4 to eq. 2

answer is 2(2)^1/2
2sqrt2