Find the point on the parabola x+y^2=0 that is closest to the point (0,-3).
the distance d is found using
d^2 = x^2 + (y+3)^2
Since x = -y^2, that means
d^2 = y^4 + (y+3)^2
Now all you do is find dd/dy = 0 and that is where the distance is minimum. Then evaluate x = -y^2 and you have your point.
Well, isn't that a quadratic question! Let's start by finding the distance between two points. We can use the good old distance formula, d = √((x2 - x1)^2 + (y2 - y1)^2). In this case, let's call the point on the parabola (x, y).
So, the distance between (0, -3) and (x, y) is d = √((x - 0)^2 + (y - (-3))^2). But we also know that the point (x, y) lies on the parabola x + y^2 = 0.
Substituting y = √(-x) into our distance equation, we have d = √(x^2 + (√(-x) - (-3))^2).
Now, let's simplify a bit. d = √(x^2 + (√(-x) + 3)^2). That's starting to look closer to something we can work with!
Now, let's crank up the clown factor and notice that to minimize the distance, we need to minimize the expression under the square root. So, we want to find the minimum value of x^2 + (√(-x) + 3)^2.
But hold on a second, we can't take the square root of a negative number! So it seems like there's no real solution here, and the closest point on the parabola to (0, -3) is indefinitely far away. It's like trying to catch a unicorn wearing a clown wig - pretty tricky!
So, in conclusion, there is no point on the parabola x + y^2 = 0 that is closest to the point (0, -3). It's like trying to find a clown without a red nose - they just don't exist!
To find the point on the parabola x + y^2 = 0 that is closest to the point (0, -3), we can use a mathematical technique called the "method of Lagrange multipliers." This method helps us find the extrema of a function subject to a constraint.
Let's define our distance function as the square of the distance between two points:
d(x, y) = (x - 0)^2 + (y + 3)^2
The constraint equation is given by the parabola:
g(x, y) = x + y^2 = 0
Now, we form the Lagrangian function:
L(x, y, λ) = d(x, y) + λ * g(x, y) = (x - 0)^2 + (y + 3)^2 + λ * (x + y^2)
To find the closest point, we need to find the values of x, y, and λ that simultaneously satisfy the following conditions:
∂L/∂x = 0 (1)
∂L/∂y = 0 (2)
∂L/∂λ = 0 (3)
Differentiating L(x, y, λ) with respect to x, y, and λ, we have:
∂L/∂x = 2x + λ = 0 (4)
∂L/∂y = 2(y + 3) + 2λy = 0 (5)
∂L/∂λ = x + y^2 = 0 (6)
Solving equations (4) and (6) simultaneously:
2x + λ = 0 (from equation 4)
x + y^2 = 0 (from equation 6)
Substituting λ = -2x into x + y^2 = 0, we get:
x + y^2 = 0
The above equation represents the parabola itself.
Now, substituting y^2 = -x into the equation x + y^2 = 0, we have:
x + (-x) = 0
Simplifying, we find:
2x = 0
So, x = 0.
Substituting x = 0 into y^2 = -x, we find:
y^2 = 0
Hence, y = 0.
Therefore, the closest point on the parabola x + y^2 = 0 to the point (0, -3) is (0, 0).
To find the point on the parabola x + y^2 = 0 that is closest to the point (0, -3), we can use the distance formula.
The distance formula states that the distance between two points (x1, y1) and (x2, y2) is given by the formula:
d = sqrt((x2-x1)^2 + (y2-y1)^2)
In this case, the point (0, -3) is given by (x1, y1) and the point on the parabola is given by (x2, y2). Let's represent the latter point as (x, y) for now.
We want to minimize the distance between these two points, so we need to minimize the distance formula expression. Since we are only interested in the positive square root, we can also minimize the square of the distance, which is equivalent.
So, the squared distance between the two points is: d^2 = (x - 0)^2 + (y - (-3))^2
Substituting the expression for y from the equation of the parabola, we get: d^2 = x^2 + (y^2 + 3)^2
Now, we need to find the point (x, y) on the parabola that minimizes this expression. To do that, we can take the derivative of d^2 with respect to x and y and set them equal to zero.
Differentiating with respect to x, we get: d^2/dx = 2x + 2y^2(dy/dx) = 0
Differentiating with respect to y, we get: d^2/dy = 2y(y^2 + 3) + 2x(dy/dy) = 0
Simplifying these equations, we have:
2x + 2y^2(dy/dx) = 0
2y(y^2 + 3) + 2x(dy/dy) = 0
Solving these two equations simultaneously will give us the values of x and y that minimize the squared distance.
Once we have the values of x and y, we can substitute them back into the equation of the parabola x + y^2 = 0 to find the coordinates of the point on the parabola that is closest to (0, -3).