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

Answer 1

Using the pythagorean theorem, the distance from (0,3) to a point (x,y) is equal to sqrt( (x-0)² + (y-3)² )

Using the parabola equation
x + y^2 = 0

Solve for x:
x = -y^2

Now substitute this into the distance formula:
sqrt( (-y²)² + (y-3)² )

Expanding out:
sqrt ( y^4 + y^2 -6y + 9 )

If this is minimized, so will the square, so just drop the sqrt.
f(x) = y^4 + y^2 - 6y + 9

Take the derivative and set it equal to zero:
f'(x) = 4y^3 + 2y - 6 = 0

Find the roots for y, calculate x and you'll have your point(s).

Answer 2

I believe there is an error in the solution presented above.

D = Distance. Minimizing the distance is equivalent to minimizing D^2.

The distance function should be D^2 = x^2 + (y+3)^2
Then the function D^2 = y^4 + y^2 + 6y + 9

Taking the derivative, simplifyin and equalizing to zero we get

4y^3 + 2y + 6 = 0

or

2y^3 + y + 3 = 0

This sum of the odd coefficients is equal to the sum of the even coefficients so y = -1 is a root.

Putting y = -1 in the x + y^2 = 0 parabola we get, x = -1.

So the point is (-1,-1).

Voila