Which point on the curve y=(x-2)^(1/2)+1 is closest to the point (4,1)?

Question

theres also the part two of the questions...what is the minimum distance?

its my summer assignment and i kind of lost quite a big amount of math formulas and knowledge since its..summer.

please help!!

thank you soo much

Answer 1

Better if you write the eqn of the curve as x = y^2 -2y +3.

Now if P(x, y) is any point on the curve then its distance from (4,1) id given by D^2 = (x-4)^2 +(y-1)^2

                   = (y^2 - 2y-1)^2 +(y-1)^2, the point satifies the eqn of the curve so value of x can replaced.

Now D is the distanc which cannot be - ve. Hence D is minimum then D^2 is minimum. Let M = D^2

so find dM/dy and decide for minimum etc.

Answer 2

The given curve y = y=√(x - 2) + 1 is a parabola, let (p, q) be the point on the curve closest to (4, 1),

therefore its distance between these 2 points will be

D = √[(p - 4)^2 + (q - 1) ^ 2]  . .... (i)

Now we want D to be minimum,

For that, first find p in terms of q, by replacing x with p, and y with q in the given curve we get,

q = √(p - 2) +1

p = (q - 1) ^ 2 + 2 . .. .(ii)

Now put this value of p in (i), and then differentiate it w.r.t. q, we will get

dD / dp = (2*(q - 1) + 4*(-2 + (-1 + q)^2)*(-1 + q))/(2*√((-2 + (-1 + q)^2)^2 + (-1 + q)^2))

Now equate the above to 0, we will get 2 values of q as,

q = 1/2 * (2 - √6) and q = 1/2 * (2 + √6)

Now substitute these in (ii), you will get,

p = 7 / 2 and p = 7 / 2 (both the values of q giving same value of p)

so now you can easily figure out that 7 / 2, 1/2 * (2 - sqrt(6)) will be the closest and distance = 1.140399165 units

But 7 / 2, 1 / 2 * (sqrt(6) + 2) = will be far from 4, 1 with distance,

2.48987745537 units.