Expert: Paul Klarreich - 2/8/2006

Question
find the shortest distance from the point(4,2) to the parabola 8x=y^2

Answer
Hi, Brian,

Your Question:  find the shortest distance from the point(4,2) to the parabola 8x=y^2

I suppose there are many ways to approach this, but here is one that looks attractive.  I am going to assume you know some calculus.

NOTE:

1. NEXT TIME PLEASE TELL ME WHAT YOU ARE STUDYING.  I NEED THE CONTEXT OF THE PROBLEM.

2. VIEW THIS IN A FIXED FONT.

At the 'closest' point, (x,y), the line from (x,y) to (4,2) would be perpendicular to the tangent line at (x,y).

Use implicit differentiation:

8 = 2y dy/dx, and solve, to get  dy/dx = 4/y, which we call m1
                                         
The slope of the line from (x,y) to (4,2) is:
    y - 2
m2 = -----
    x - 4

Now if two lines are perpendicular the product of their slopes is -1, so we have m1m2 = -1:
4  y - 2
--- ----- = -1
y  x - 4

4y - 8 = -y(x - 4)
4y - 8 = -xy + 4y
xy = 8

Now one more fact:  (x,y) is on our parabola.  So x = y^2/8

y^2
--- y = 8
8

y^3 = 64

So y = 4, and x = 2.