Make all lines at a point a on the curve y = x * 2 (x is greater than or equal to 0) so that the area enclosed by the curve and X axis is 1 / 12. Try to find the tangent point a and tangent square What's the process? I'll understand. Help

Make all lines at a point a on the curve y = x * 2 (x is greater than or equal to 0) so that the area enclosed by the curve and X axis is 1 / 12. Try to find the tangent point a and tangent square What's the process? I'll understand. Help

Let the tangent point be a (a, a ^ 2), then the derivative of y = x ^ 2 at a is 2a, and the tangent equation is Y-A ^ 2 = 2A (x-a)
The area of ABC is 1 / 2 * ac * BC = 1 / 2 * a ^ 2 * A / 2 = 1 / 4A ^ 3
Let the origin be o, and it is easy to know from the knowledge of definite integral that the area enclosed by the line OC AC and the curve Ao is 1 / 3A ^ 3
According to your conditions:
1/3a^3-1/4a^3=1/12
The solution is: a = 1
So the coordinates of point a are (1,1)
It is also easy to get the tangent equation: y = 2x-1