As shown in the figure, it is known that the line y = KX + B intersects the parabola y = x2 ^ with P and Q, the abscissa of P is 2 and intersects the X axis with m (2,0) to find the expression of the line y = KX + B Find the area of the triangle formed by the point P, Q and the origin The abscissa of P is - 2

As shown in the figure, it is known that the line y = KX + B intersects the parabola y = x2 ^ with P and Q, the abscissa of P is 2 and intersects the X axis with m (2,0) to find the expression of the line y = KX + B Find the area of the triangle formed by the point P, Q and the origin The abscissa of P is - 2

1. Because P is on the parabola y = x & # 178; and the abscissa is - 2, so the coordinate (- 2,4) P (- 2,4), m (2,0) is substituted into the linear equation y = KX + b-2k + B = 42K + B = 0, and the solution is k = - 1, B = 2, so the linear equation is y = - x + 22, y = - x + 2, y = x & # 178; the simultaneous solution is x = - 2, y = 4 or x = 1, y = 1, q (1,1) y = - x + 2

As shown in the figure, the straight line y = KX + B passing through the point F (0,1) intersects with the parabola y = 14x2 at two points m (x1, Y1) and n (X2, Y2) (where X1 < 0, X2 > 0). (1) find the value of B; (2) find the value of x1 · x2; (3) make the perpendicular of the straight line L: y = - 1 through M and N, and the perpendicular feet are M1 and N1 respectively Whether the circle of is tangent. Please explain the reason

(1) ∵ the line y = KX + B passes through the point F (0, 1), ∵ B = 1; (2) ∵ the line y = KX + B intersects the parabola y = 14x2 at two points m (x1, Y1) and n (X2, Y2), ∵ it can be concluded that: KX + B = 14x2, sorted out: 14x2-kx-1 = 0, ∵ a = 14b = - K, C = - 1 ∵ x1 · x2 = CA = - 4; (3) ① △ m1fn1 is a right triangle (point F is the right vertex). The reasons are as follows: let the intersection of line L and Y axis be F1, FM In this paper, m1n112 = (x1-x2) 2 = X12 + 2 + 4, m1n112 = (x1-x2) 2 = (x1-x2) 2 = (x1-x2) 2 = X12 + x22-2x12-2x12-2x12 = X12 + X22 + 8, {FM12 + fn12 = m1n112, and {△ m1fn1 is the right triangle with F point as the right vertex. ② y = - 1 and the circle with Mn diameter is tangent to the circle with Mn diameter. The reasons are as follows: for the following: for the following reasons: for the following reasons: for the following: for the following reasons: for the following reasons: for the following reasons: for the following reasons: Mn2 = MH2 + NH2 = 2 = MH2 + NH2 = (MH2 + NH2 = 2 = 2 + NH2 = (x1-2 = (x1-x2) 2 + NH2 = (x1-x2) 2) 2) 2 = (x1-x2) 2) 2 + (x1-2) 2 + (x1 K2 (x1) -X2) 2, = (K2 + 1) (x1-x2) 2, = (K2 + 1) [(x1 + x2) 2-4x1 · x2] = (K2 + 1) (16k2 + 16) = 16 (K2 + 1) 2, ｜ Mn = 4 (K2 + 1), take the midpoint P, P1, PP1 = 12 (MM1 + NN1) = 12 (Y1 + 1 + Y2 + 1) = 12 (Y1 + Y2) + 1 = 12K (x1 + x2) + 2 = 2k2 + 2, ｜ PP1 = 12mn of Mn and m1n1 respectively, that is, the distance from the midpoint of Mn to the straight line L is equal to half of the length of Mn Tangent to L

It is known that a (x1, Y1), B (X2, Y2), C (X3, Y3) are three points on the parabola y ^ 2 = 2px (P > 0),
The distances AF, BF and CF from them to the focus f form an arithmetic sequence, and the proof is: 2Y2 ^ 2 = Y1 ^ 2 + Y3 ^ 2

From 2BF = AF + CF
According to the definition of parabola, AF = X1 + P / 2, BF = x2 + P / 2, CF = X3 + P / 2
Yide 2x2 = X1 + x3
And Y ^ 2 = 2px
So 2Y2 ^ 2 = Y1 ^ 2 + Y3 ^ 2

The parabola y = x-kx + K + 4 intersects Y-axis at point C, intersects a and B with x-axis, and the abscissa is an integer
Find the parabola and its vertex

We know that the equation x ^ 2-kx + K + 4 = 0 has two unequal real roots, and they are all integers
Discriminant > 0
k^2-4(k+4)>0
k^2-4k-16>0
(k-2)^2>20
k> 2 (1 + √ 5) or K