The point on the curve?

Question

The point on the curve x^2+2y=0 that is nearest the point (0, -1/2) occurs where y is?

 

Answer: 0

Answer 1

Hi Sun,

 

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)

 

                    or

 

    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.

Answer 2

What is the y-value of the point on the curve x^2+2y=0 that is nearest the point (0, -1/2)?

Let’s find the distance between (0, -1/2) and a generalized point on the curve, (x,-x^2/2).

Draw a right triangle with horizontal and vertical legs. Label the end points of the hypotenuse (0, -1/2) and (x,-x^2/2). The distance, D, between the two points is the length of the hypotenuse, which we can get from the Pythagorean Theorem.

D^2 = (0-x)^2 + (-1/2 - -x^2/2)^2
D^2 = x^2 + (-1/2 + x^2/2)^2
D^2 = x^2 + ((x^2-1)/2)^2
D^2 = x^2 + ((x^2-1)^2)/4
4D^2 = 4x^2 + (x^2-1)^2
4D^2 = 4x^2 + x^4 - 2x^2 + 1
4D^2 = x^4 + 2x^2 + 1
4D^2 = (x^2 + 1)^2
D^2 = ((x^2 + 1)^2)/2^2
D = (x^2 + 1)/2
D = x^2/2 + 1/2, which is a parabola opening up with minimum when x = 0.

Substituting x = 0 into x^2+2y=0 => y = 0.

Answer 3

Goal: Minimize D=√[x²+(y-1/2)²]

Constraint: G=x²+2y=0

 

Let's minimize S=D² instead. We will use Lagrange Multipliers.

 

S = x²+y²-y+1/4

 

We need ∇S=λ∇G

 

We get the system

 

(i) 2x = 2λx

(ii) 2y - 1 = 2λ

(iii) x² + 2y = 0

 

Equation (i) is true if either x = 0 or λ = 1

 

If x = 0 then we have y = 0 for the point (0,0)

 

If λ = 1 then solving (ii) gives y = 1/2, and then there are no real solutions for x in (iii).

 

Hence the minimum is at (0,0) making y=0.