Make all lines at a point a on the curve y = X2 (x ≥ 0) so that the area between them and the figure surrounded by the curve and x-axis is 112. Try to find the coordinates of tangent point a and the tangent equation passing through tangent point a

As shown in the figure, let the tangent point a (x0, Y0) and y '= 2x, the tangent equation passing through point a is y-y0 = 2x0 (x-x0), i.e. y = 2x0 x-x02. Let y = 0, then x = X02, i.e. C (X02, 0). Let the area of the figure enclosed by the curve, the tangent line passing through point a and the x-axis be s.s-curved triangle AOB = ∫ x000x2dx = 13x3 | X00 = 13x03, s △ ABC = 12 | BC | · | ab | = 12 (x0-x02) · X02 = 14x03 3 = 112. X0 = 1, so the coordinate of tangent point a is (1,1), and the tangent equation is y = 2x-1

Make a tangent line at a point a [a, a ^ 2] on the curve y = x ^ 2 [x is greater than or equal to zero], so that the area enclosed by the curve and X axis is 1 / 12
Advanced mathematics try to calculate the coordinates of tangent point a (a, a ^ 2) thank you

In addition, the intersection of tangent and x-axis is a / 2, the area is the square of integral 0 → ax, plus a / 2 as the base height and the square of a multiplied by half is 12. The cube of a is one, so the coordinate of a is (1.1)

Make all lines at a point a on the curve y = X3 (x ≥ 0) so that the area of the figure enclosed by the curve and X axis is 1 / 12. Try to find the tangent equation of the tangent point a

Y = x ^ 3 and y = 3x ^ 2
Let a be (x1, X1 ^ 3)
The tangent is y = 3x1 ^ 2 (x-x1) + X1 ^ 3
And ∫ (x ^ 3-3x1 ^ 2 (x-x1) - X1 ^ 3) DX (0 ~ x1) = 1 / 12
1/4x1^4-3/2x1^4+2x1^4=1/12
We get X1 = √ 3 / 3
The tangent equation is y = X-2 √ 3 / 9

Make all lines at a point a on the curve y = X2 (x ≥ 0) so that the area between them and the figure surrounded by the curve and x-axis is 112. Try to find the coordinates of tangent point a and the tangent equation passing through tangent point a