S.O.S. Mathematics CyberBoard

the point M is on the parabola x^2=64y.Which is the minimum distance between M and the line 3x+4y+37=0?

I would do it like this...

M is on x^2=64y, or y=(x/8)^2. So, M coordinates have the form (x, (x/8)^2).

Distance from M to given line is:

D = |3x + 4(x/8)^2 + 37| / sqrt (9 + 16)

D^2 = (3x + x^2/16 + 37)^2 / 25 (let's call D^2 = W)

If D is minimum, then D^2 = W is also minimum (remember that D>=0).

dW/dx = (1/25) (2(3x + x^2/16 + 37) (3 + x/8)

W is minimum when dW/dx = 0:

(1/25) (2(3x + x^2/16 + 37) (3 + x/8) = 0

So, either

(3x + x^2/16 + 37) = 0 -> no real roots

or

(3 + x/8) = 0 -> x = -24

Putting x = -24 in the coordinates of M, we have the answer

M(-24, 9)

_________________
Vini®

Good job,i am checking it on the paper now.Thx!