Ask a question

The point on the curve?

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

3 Answers by Expert Tutors

Tutors, sign in to answer this question.
Andrew K. | Expert Math and Physics Tutor - Many successful students!Expert Math and Physics Tutor - Many suc...
5.0 5.0 (195 lesson ratings) (195)
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)
    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.
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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.
Roman C. | Masters of Education Graduate with Mathematics ExpertiseMasters of Education Graduate with Mathe...
4.9 4.9 (363 lesson ratings) (363)
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.