The parabola y = - 3 / 4x ^ 2 + 3 intersects the x-axis at points a and B, the straight line y = - 3 / 4x + B intersects at points B and C, and the straight line y = - 3 / 4x + B intersects the y-axis at points E (1) Write the analytical expression of the straight line BC (2) Finding the area of △ ABC (3) If point m moves from a to B at the speed of one unit length per second on line AB (not coincident with a and b), and point n moves from B to C at the speed of two units length per second on ray BC. Let the motion time be T seconds, write out the functional relationship between the area s of △ MNB and T, and find out the maximum area of △ MNB when point m moves for how long?

The parabola y = - 3 / 4x ^ 2 + 3 intersects the x-axis at points a and B, the straight line y = - 3 / 4x + B intersects at points B and C, and the straight line y = - 3 / 4x + B intersects the y-axis at points E (1) Write the analytical expression of the straight line BC (2) Finding the area of △ ABC (3) If point m moves from a to B at the speed of one unit length per second on line AB (not coincident with a and b), and point n moves from B to C at the speed of two units length per second on ray BC. Let the motion time be T seconds, write out the functional relationship between the area s of △ MNB and T, and find out the maximum area of △ MNB when point m moves for how long?


As shown in the picture



Given that the parabola y ^ 2 = 4x, the line L passing through the point m (0,2) intersects the parabola at two points a and B, and the line L intersects the X axis at point C


1) You first use the similar shape to describe the following: XC / XB = XC / XB = XA / XC a (XA, ya) B (XB, Yb) B (XB, Yb) and then set the linear equation y = K (x-xc) this way, you have to explain that the slope does not exist, the situation can't intersect with AB, and then you can use the similar shape to describe the following: first, you use the similar shape to describe the following: XC / XB = XC / XB = XC / XB = XC / XB = XA / XB = XA / XB = XA / XB = XA / XB = XA = 4x = 4x = 4x = 4x = 4x = 4x = 4x = 4x = 4x = 4x = 4x = 4x = 4x = K / 4Y ^ 2 / 4Y ^ 2 / 4Y ^ 2 / 4Y ^ 2-y ^ 2-y ^ 2-y ^ 2-y ^ 2-y ^ 2-y it is proved that the vector Ma = α vector AC, Vector MB = β vector BC MB vector = MC vector + CB vector note that the vector is a vector, and the vector is a vector