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.
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.