Question

Find the point (x,) on the graph of the curve y=√4x+13 that is the closest to the fixed point (5,0).

Answer 1

EDIT: Looking for the point (x,y) not (x,). My apologies for the typo.

Answer 2

is the equation
y = √(4x + 13) , or
y = √(4x) + 13

I will assume y = √(4x+13)

let the point of contact be P(x,y)
or P(x, √(4x+13) )

let D be the distance from P to (5,0)
D = (√(4x+13) - 0)^2 + (5-x)^2 )
= √( (4x+13) + (5-x)^2 )
square both sides to make it easier to differentiate
D^2 = 4x+13 + (5-x)^2
= 4x + 13 + 25 - 10x + x^2
= x^2 - 6x + 38

2D dD/dx = 2x - 6

for a max/min of D , dD/dx = 0

2x - 6 = 0
x = 3
so P is (3, √25) or P(3,5)

Answer 3

just noticed that in my
D = (√(4x+13) - 0)^2 + (5-x)^2 )
it should have been
D = (√(4x+13) - 0)^2 + (x-5)^2 )
as well as the following 2 lines.

but since we squared it, that error did not matter.