Question

Find the minimum distance from the parabola
x + y^2 = 0
to the point (0,-3).

Answer 1

make a sketch
At P(x,y) , the point of shortest distance the slope of the line from P to (0-3) must be perpendicular to the slope of the tangent at P

x + y^2 = 0
1 + 2y dy/dx = 0
dy/dx = -1/2y

slope of line from (0,-3) to P = (y+3)/x

then (y+3)/x = 2y/1 , (slopes must be negative reciprocals)
y+3= x(2y)
but x = -y^2
y+3 = -2y^3
2y^3 + y + 3 =0
easy to see that y = -1 is a solution
then x = -y^2 = -1
P is (-1,-1)
and distance = √( (-1)^2 + (-3+1)^2 ) = √5

Answer 2

But a different way, same answer :)