Find the point on the parabola x+y^2=0 that is closest to the point (0,-3)

Let us say the coordinates of the point closest to (0,-3) are (a,b).
The distance (L) between these two points can be given as;
L = sqrt((0-a)^2+(-3-b)^2)
By getting L^2
L^2 = a^2+(3+b)^2 -----(1)
 
However (a,b) is on the parabola. So we can say;
a+b^2 = 0
a = -b^2
By substituting a = -b^2 on equation (1)
L^2 = b^4+(3+b)^2
L^2 = b^4+b^2+6b+9 ----(2)
 
The maximum/minimum of L is given when dL/db = 0
By first derivative on (2)
2LdL/db = 4b^3+2b+6
For maximum and minimum dL/db = 0
4b^3+2b+6 = 0
 
In these complex cases it is better to apply b = -1, b = +1 or b = 0 and see whether it solves equation.
You can see at b = -1 gives you one solution. 
So we can write,
(b+1)(pb^2+qb+r) =4b^3+2b+6 where b^2 is not equal to 0.
pb^3+(q+p)b^2+(r+q)b+r =4b^3+2b+6
By comparing components,
p = 4
r = 6
q = -4
 
pb^2+qb+r = 4b^2-4b+6
4b^2-4b+6 = 0
b^2-2b+3 = 0
delta = (-2)^2-4xx1xx3<0
So this part does not have real solutions. They have complex solutions.
 
When b<-1 say b = -2 then
 dL/db = (4b^3+2b+6)/(2L) = (-32-4+6)/(2L)
Since L>0 then dL/db<0.
 
When b>-1 say b = 0 then
 dL/db = (4b^3+2b+6)/(2L) = (0+0+6)/(2L)
Since L>0 then dL/db>0.
 
Since at b= -1 the parabola changes from negative gradient to a positive gradient, that means the curve has a minimum. So the length between points is minimum.
When b = -1 then;
a = -b^2
a = -1
 
So the closest point to (0,-3) is (-1,-1) which is on x+y^2 = 0 parabola.
 
http://clas.sa.ucsb.edu/staff/lee/Max%20and%20Min's.htm

http://www.mesacc.edu/~marfv02121/readings/nearest_point/index.html

https://en.wikipedia.org/wiki/Quadratic_equation