Let the intersection points of the parabola y = 4-x ^ 2 and the straight line y = 3x be a and B. point P moves from a to B on the arc of the parabola, 1. Find the coordinates (a, b) of point P to maximize the area of triangle PAB 2. Prove that the figure enclosed by parabola y = 4-x ^ 2 and straight line y = 3x is divided into two parts with equal area by straight line x = a

Let the intersection points of the parabola y = 4-x ^ 2 and the straight line y = 3x be a and B. point P moves from a to B on the arc of the parabola, 1. Find the coordinates (a, b) of point P to maximize the area of triangle PAB 2. Prove that the figure enclosed by parabola y = 4-x ^ 2 and straight line y = 3x is divided into two parts with equal area by straight line x = a

A. The coordinates of two points B are (- 4, - 12), (1,3) respectively
And - 4

The parabola y = 3x & # 178 intersects with the straight line y = - 5x + 2 at two points a and B to find the area of the triangle OAB

Two simultaneous equations, 3x & # 178; = - 5x + 2, so x = - 2 or x = 1 / 3, so a (- 2,12), B (1 / 3,1 / 3), and the intersection of y = - 5x + 2 and Y axis is C (0,2), so the area of △ AOB is △ AOC + △ BOC = 2 * 1 / 3 divided by 2 + 2 * 2 divided by 2 = 7 / 3, so the area is equal to 7 / 3

Find the area of a curved triangle enclosed by parabola y = x & #, straight line x = 1 and X

A mathematical problem about parabola
It is known that the line intersecting the focus of the parabola y2 = 2px (P greater than 0) is at two points AB, and ab = 5 / 2p. The AB equation of the line is solved

If the line AB is perpendicular to the x-axis, then x = P / 2Y ^ 2 = P ^ 2, y = P, y = - P. at this time, ab = P - (- P) = 2p is not true, so the line AB is not perpendicular to the x-axis, so the slope of the line AB exists y-0 = K (X-P / 2) = KX KP / 2Y ^ 2 = 2pxk ^ 2x ^ 2-k ^ 2px + K ^ 2p ^ 2 / 4 = 2pxk ^ 2x ^ 2 - (k ^ 2p + 2P) x + K ^ 2p ^ 2