find the shortest distance from point(4,0) to a point on the parabola y^2=2x?

Answer 1

Let say shortest distance from (4,0) to point (x1,y1) at y^2=2x parabola is s

s^2 = (x1 - 4)^2 + (y1 - 0)^2
s^2 = (x1 - 4)^2 + y1^2
from parabola y1^2 = 2x1,
s^2 = (x1 - 4)^2 + (2x1)
s^2 = x1^2 - 8x1 + 16 + 2x1
s^2 = x1^2 - 6x1 + 16
differentiate in respect with x1
2s ds/dx1 = 2x1 - 6
To get minimum s , s = 0
2x1 - 6 = 0
x1 = 3
y1 = +/- sqrt(6)

minumum distance
= sqrt[(3 - 4)^2 + sqrt(6)^2]
= sqrt[1 + 6]
= sqrt(7)

Answer 2

We can write a general point on the curve as (y^2/2, y). (We do it this way around so we don't have to choose between poaitive and negative square roots.) Let f(y) be the square of the distance from this point to (4, 0), i.e. f(y) = (4 - y^2 / 2)^2 + (0-y)^2
<=> f(y) = 16 - 4y^2 + y^4/4 + y^2 = 16 - 3y^2 + y^4/4.

This is a quadratic in y^2, so we know the minimum value will occur at y^2 = -(-3) / 2(1/4) = 6. Then f(y) = 16 - 3(6) + 6^2 / 4 = 7. So the minimum distance is √7.