Find the point on the curve y^2=4x closest to the point (8,0).

Question

My Work: Using the Distance formula-
my answer i got was (7.821,5.593) but idk if i did it right...

Answer 1

One way is to find the length of the line normal to the curve which passes through the point.

The tangent at any point has slope 1/√x, so the normal line has slope -√x.

The line through (8,0) and (x,2√x) has slope 2√x/(x-8), which we want to be -√x.

2√x/(x-8) = -√x
-2 = x-8
x = 6

The point is (6,2√6)

Or, using the distance formula,

d = √((x-8)^2+(y)^2)
= √(((x-8)^2+4x)
= √(x^2-12x+64)

d' = (x-6)/√(x^2-20x+64)
minimum distance d is where
d'=0 at x=6, as above

Answer 2

Ignore the comment about the length of the normal line. The question did not actually ask for the distance, just the point on the curve.

We just wanted to find the point where the normal to the curve passes through (8,0)

Answer 3

Oh so then i plug x=6 back to the original to get Y