Consider the graph of y=9-x^2. At which point(s) on the graph will the normal line to the curve pass through the point (1,4)?

Let (x1,y1) be the point on the curve. Then the y1 =
9-x1^2.

The slope of the curve at x1 is dy/dx at x=
x1.

At x= x1, y1 =
9-x1^2.

Therefore dy/dx = (9-x^2)' =
-2x.

Therefore the slope m of the normal at x1 is given by
m = {-1/(dy/dx) at x = x1} = - 1/(-2x1) =
1/2x1

Therefore the equation of the normal at x = x1 is
given by:

y-y1 = m(x-x1).

y -
(9-x1^2) = (1/2x1)(x-x1)...(1).

Since the normal at (1)
passes through the point (1,4), the ccordinates of the point (1,4) should satisfy the
equation of the line at (1):

4 - (9-x1^2)  =
(1/2x1)(1-x1).

(-5 +x1^2)2x1 =
1-x1.

2x1^3 -10x1+ x1 - 1 =
0.

2x1^3 - 9x1 - 1 =
0.

Therefore x1 = 2.2 nearly 4.4 nearly.Or the normal at
(2.2 , 4.2)

passes through (1,4).