If the line passing through the focus F of the parabola y2 = 4x intersects with the parabola at two points a and B, then OA · ob=______ ．

If the line passing through the focus F of the parabola y2 = 4x intersects with the parabola at two points a and B, then OA · ob=______ ．

The focus coordinate of the parabolic y2 = 4x is (1,0), (the equation of AB is y = K (x-1), the equation of AB is y = K (x-1), and the equation is y = K (x-1) from y2 = 4xxy = K (x − 1) to get k2x2 - (2k2 + 2 + 4) x + K2 = 0, let a (x1, Y1, Y1), B (X2, Y2), then X1 + x2 = 2k2 + & nbsp; 4k2, the equation of AB AB is y = K (x-k (x-1), the equation equation of AB AB ab AB is y = K (x-1 (x-1) for the equation equation of AB AB AB AB AB AB is y = 1, the equation equation of AB AB AB AB (y 1-1, y1-1 \\\ \\\11\\& nbsp; = 3, so the answer is: - 3

It is known that the focus of parabola C: y2 = 4x is f
(1) Point a, P satisfies
AP
=-2
FA
When point a moves on parabola C, the trajectory equation of moving point P is obtained;
(2) Is there a point Q on the x-axis so that the symmetric point of the point Q with respect to the straight line y = 2x is on the parabola C? If so, find the coordinates of all the points Q satisfying the condition; if not, explain the reason

The focus of parabola C: y ^ 2 = 4x is f (1,0),
Let a (m, n), P (x, y) be the points, and the vector AP = - 2fa is the vector
(x-m,y-n)=-2(m-1,n),
∴x-m=-2(m-1),y-n=-2n,
∴m=2-x,n=-y.
Point a is on the parabola C,
The equation (- y) ^ 2 = 4 (2-x), that is, y ^ 2 = - 4 (X-2), is the trajectory equation of the moving point P
(2) Let Q (Q, 0), then q satisfies R (4T ^ 2,4t) with respect to the symmetric point R (4T ^ 2,4t) of the line y = 2x
The slope of QR = 4T / (4T ^ 2-Q) = - 1 / 2, 8t = q-4t ^ 2, ①
The midpoint of QR ((Q + 4T ^ 2) / 2,2t) is on the straight line y = 2x, that is, 2T = q + 4T ^ 2, ②
② - 1, - 6T = 8t ^ 2, t = 0 or - 3 / 4,
Substituting ①, q = 0 (rounding), or - 15 / 4
∴Q(-15/4,0).

If | ab | = 10, then the distance from the midpoint P of AB to the Y axis is equal to______ ．

Parabola y2 = 4x, focal point E (1, 0), collimator L: x = - 1, because the midpoint of AB is p, passing through & nbsp; a, P, B to make the vertical line of collimator respectively, the vertical feet are C, F, D, PF intersects the longitudinal axis at point h, as shown in the figure: pf = AC + BD2 = AE + Eb2 = AB2 = 5, ∧ pH = pf-fh = 5-1 = 4, so the answer is: 4

If | ab | = 10, then the distance from the midpoint of AB to the Y axis is equal to ()
A. 1B. 2C. 3D. 4

Parabola y2 = 4x focal point (1, 0), the collimator is l: x = - 1, let the midpoint of AB be e, pass through a, e, B to make the vertical line of the collimator respectively, the vertical feet are C, G, D, EF intersects the longitudinal axis at point h, as shown in the figure: then we can know from the median line of right angled trapezoid, EF = AC + BD2 = AF + FB2 = AB2 = 5,... Eh = EG-1 = 4