Question

What point on the parabola y=1−x2 is closest to the point (1,1)? The point on the parabola y=1−x2 that is closest to the point (1,1) is (Type an ordered pair. Round to two decimal places as needed

Answer 1

To find the point on the parabola y=1−x^2 that is closest to the point (1,1), we need to find the minimum distance between these two points. By taking the derivative of the distance function and setting it equal to zero, we can find the x-coordinate of the point on the parabola that is closest to (1,1).

Step-by-step explanation:

To find the point on the parabola y=1−x^2 that is closest to the point (1,1), we need to find the minimum distance between these two points.

The distance between two points (x1, y1) and (x2, y2) can be found using the distance formula: d = sqrt((x2-x1)^2 + (y2-y1)^2).

In this case, the distance between (x, 1−x^2) and (1, 1) is given by d = sqrt((1-x)^2 + (1-(1-x^2))^2).

To find the point on the parabola that minimizes this distance, we can take the derivative of this distance function with respect to x, set it equal to zero, and solve for x.

By solving this equation, we can find the x-coordinate of the point on the parabola that is closest to (1,1). Once we have this x-coordinate, we can substitute it back into the equation y=1−x^2 to find the corresponding y-coordinate.

Answer 2

The closest point on the parabola y=1-x² to the point (1,1) is found by minimizing the square of the distance between a generic point on the parabola and (1,1). This involves calculating the derivative of the squared distance function and determining its critical points by setting the derivative equal to zero. Solving for x gives the x-coordinate of the closest point.

Step-by-step explanation:

To find the point on the parabola y=1-x² that is closest to the point (1,1), we can employ the method of minimizing the square of the distance between a generic point (x, 1-x²) on the parabola and the point (1,1).

To do this, let's first set up the squared distance function D²:

D² = (x-1)² + (1-x²-1)²

Expanding and simplifying this, we have:

D² = x² - 2x + 1 + x´ - 2x³ + 1

To find the minimum distance, we need to differentiate this with respect to x and set the derivative equal to zero:

D' = 2x + 4x³ - 6x²

The critical points can be found by setting D' equal to 0:

0 = 2x(1 + 2x² - 3x)

We solve this equation for x to find the x-coordinate of the closest point on the parabola.