Let's first rearrange to give an expression for y in terms of x:
x2 + 2y = 0
Subtracting x2 and dividing by 2:
y = (-x2)/2
If we plotted this, it would be a downward-opening parabola, centered on the y-axis.
Now, let's come up with an expression that would calculate the distance between a point on that parabola and any other point on the graph. The pythagorean theorem shows us how to calculate the length of the hypotenuse of a right triangle, and that's exactly what we would be dealing with here, if we treat the (difference in the y values) and (difference in x values) as the two perpendicular sides of a triangle.
So, we could say that the actual (hypotenuse) distance between two points would be √((y1-y2)2 + (x1-x2)2)
Let's treat x1 and y1 as the point along the parabola, and x2 and y2 as the (0, -1/2) point that is given. We want to determine the x1 and y1 points, so we will leave them as variables in our distance expression:
distance = √((y - (-1/2))2 + (x - 0)2)
Now, since we have an expression for y in terms of x, let's replace y with that expression, so that we will have only one variable in our equation:
distance = √((-x2/2 + 1/2)2 + x2)
distance = ((-x2/2 + 1/2)2 + x2)1/2
Let's expand the first squared term, using FOIL:
distance = ((x4/4 - x2/2 + 1/4) + x2)1/2
Combining the x2 terms:
distance = ((x4/4 + x2/2 + 1/4)1/2
Knowing that we want to find the point where this distance is a minimum, just from examining the equation, the smallest we could ever make this distance would be (1/4)1/2 = 1/2, if x = 0. Because of the fact that all of the x exponents are even, any other x-value that we could possibly pick, positive or negative, would only make the distance larger.
So, we know that the distance is minimized when x=0. Since we want to know what the associated y-value is, we plug in 0 for x in the quadratic equation we derived from the given expression.
y = -(0)2/2 = 0
This also checks with what we logically concluded to be the minimum possible distance: the distance between the points (0,0) and (0,-1/2) is 1/2.