Q.133. Applications of derivatives, Maxima/Minima

Question 133.
A point P (x,x^2) lies on the curve y=x^2. Calculate the minimum distance from point A (2, -1/2) to point P.

Answer 133.
Shortest distance will be along a normal to the curve through A.
y = x^2 => dy/dx = 2x
Slope of normal = - 1/2x
=> - 1/2x = (x^2 + 1/2) /(x - 2)
=> - (x - 2) = 2x^3 + x
=> 2x^3 + 2x - 2 = 0
=> x^3 + x - 1 = 0
 Using Wolfram Alpha (Link below),
x ≈ 0.682328 and x^2 = 0.465571

=> Minimum distance
= √[(2 - 0.682328)^2 + (-1/2 - 0.465571)^2]
= 1.633581.

For confirmation of the answer obtained analytically as above, refer to the global minimum in the second link of Wolfram Alpha as under.

Alternate Method:
AP is minimum when AP^2 is minimum
AP^2
= (x - 2)^2 + (x^2 + 1/2)^2
= x^4 + 2x^2 - 4x + 17/4
d/dx (AP^2) = 0
=> 4x^3 + 4x - 4 = 0
=> x^3 + x - 1 = 0
This equation is the same as I got by a different approach as above and its solution can be obtained using Wolfram Alpha as above.

LINK 1 to WolframAlpha: Giving solution of the equation x^3 + x - 1.
LINK 2 to WolframAlpha: Giving the minimum distance with a graph.
 
LINK to YA!