Find the x–coordinate of the point on the parabola y=x^1/2 nearest to the point (4, 0).
notice the point (4,0) does not lie ON the curve.
So your sketch should show that.
let that point be (a,b)
so we know b = √a or a = b^2
dy/dx = (1/2)x^(-1/2) or 1/(2√x)
so at the point (a,b) the slope of the tangent is 1/(2√a)
I will use the property that at closest point the tangent would be perpendicular to the line joining (a,b) to (4,0)
slope of line from (a,b) to (4,0) = b/(a-4)
then 1/(2√a) = (a-4)/-b
2√a(a-4) = -b
but remember that b = √a , so
2√a(a-4) = -√a
2(a-4) = -1
2a - 8 = -1
a = 7/2 , then b = √7/√2 or √14,2 after rationalizing it
the closest point is (7/2 , √14/2 )