Find a point on the parabola?

Question

y= 9 - x^2 closet to the point (3,9)

Answer 1

Write out the equation for distance between these points:

x difference = 3 - x
y difference = 9 - (9-x²) = x²

We want to minimize:

d = √ ( xdiff² + ydiff² )

However, we can minimize:

d² = xdiff² + ydiff²

instead, to avoid the square root.

d² = (3 - x)² + x^4 = 9 - 6x + x² + x^4

Now we want to find critical values of this, so we differentiate the expression:

x^4 + x² - 6x + 9

4x³ + 2x - 6 = 0

2 (x - 1) (2x² + 2x + 3) = 0

The quadratic factor does not have real roots, so the only solution is x=1 gives us y = 8.

Thus the closest point is: (1,8)

Answer 2

Let (x,y) be the point on the parabola closest to (3,9),
then
y = 9 - x^2
and the point to find is (x, 9 - x^2)
the square of distance between (x,y) & (3, 9) is
E = (x - 3)^2 + (y - 9)^2
= (x - 3)^2 + (9 - x^2 - 9)^2
= (x - 3)^2 + x^4

To minimize E, dE/dx =0
dE/dx = 2(x -3) + 4x^3 = 0
4 x^3 + 2x - 3 = 0
x = 0.727
y = 9 - 0.727^2 = 8.47
is the point!

Answer 3

s^2 = (x - 3)^2 + (9 - x^2 - 9)^2
s^2 = x^2 - 6x + 9 + x^4
4x^3 + 2x - 6 = 0
2x^3 + x - 3 = 0
x = 1
(x,y) = (1,8)