# 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.