At what point on the parabola y = 1-x^2 does the tangent have the property that it cuts from the first quadrant a triangle of minimum area? Find the minimum area.

To find the point on the parabola where the tangent cuts a triangle of minimum area in the first quadrant, we need to determine the minimum area triangle formed between the tangent and the x-axis.

First, let's find the equation of the tangent line to the parabola at a general point (x, y). The slope of the tangent line is the derivative of the curve at that point. Taking the derivative of y = 1 - x^2 with respect to x:

dy/dx = -2x

Since the slope of the tangent line is equal to dy/dx at the point of tangency, the equation of the tangent line becomes:

y - (1 - x^2) = -2x(x - x_0)

where (x_0, y_0) is the point of tangency. Rearranging this equation gives:

y = -2x^2 + 2xx_0 - x_0^2 + 1

Now, we want to find the value of x_0 such that the triangle formed by the tangent line and the x-axis has the minimum area. The area of a triangle is given by A = (1/2)bh, where b represents the length of the base and h represents the height.

In this case, the base of the triangle is the distance between the x-axis and the point of intersection of the tangent line with the x-axis. Since the tangent line intersects the x-axis when y = 0, we can set y = 0 in the equation of the tangent line to find the x-coordinate of the intersection point:

-2x^2 + 2xx_0 - x_0^2 + 1 = 0

Simplifying this equation gives:

2x^2 - 2xx_0 + x_0^2 - 1 = 0

Using the quadratic formula, we can solve for x:

x = (2xx_0 ± √(4x_0^2 - 4(2)(x_0^2 - 1)))/(4)

Simplifying further gives:

x = (x_0 ± √(4 - 4x_0^2))/(2)

From here, we can consider two cases:

Case 1: Positive value of x

To find the root with the positive value of x, we take the positive square root of the discriminant:

x = (x_0 + √(4 - 4x_0^2))/2

Case 2: Negative value of x

To find the root with the negative value of x, we take the negative square root of the discriminant:

x = (x_0 - √(4 - 4x_0^2))/2

Now, compute the height of the triangle. Since the height is the y-coordinate of the intersection point between the tangent line and the x-axis, we substitute x into the equation of the tangent line:

y = -2x^2 + 2xx_0 - x_0^2 + 1

Once you have the values of x and y, calculate the area A = (1/2)bh using the base (x) and height (y).

Next, we need to minimize the area of the triangle with respect to x_0. To do this, take the derivative of A with respect to x_0 and set it equal to zero. Solve this equation to find the value of x_0 that gives the minimum area.

Finally, substitute the value of x_0 into the equation for x and y to get the coordinates of the point on the parabola where the tangent cuts the triangle of minimum area. Substitute these values into the area formula to find the minimum area.