The center F of the circle M: x = 1 + cos θ is a parabola e: x = 2pt & sup2; the focal point of the circle M: x = 1 + cos θ is a straight line intersection parabola e passing through the focal point F at two points y = sin θ y = 2pt Find the value range of afbf The center F of the circle M: x = 1 + cos θ y = sin θ is a parabola e: x = 2pt & sup2; the focal point of y = 2pt passes through the straight line intersection parabola e of the focal point F, and the value range of AF · BF is calculated at two points ab

The center F of the circle M: x = 1 + cos θ is a parabola e: x = 2pt & sup2; the focal point of the circle M: x = 1 + cos θ is a straight line intersection parabola e passing through the focal point F at two points y = sin θ y = 2pt Find the value range of afbf The center F of the circle M: x = 1 + cos θ y = sin θ is a parabola e: x = 2pt & sup2; the focal point of y = 2pt passes through the straight line intersection parabola e of the focal point F, and the value range of AF · BF is calculated at two points ab

(4, infinity)
AF.BF=4/ (tana)^2

What is the shortest distance from the point of parabola y = x & # to the line 2x-y-4 = 0

Let P (x0, Y0) be a point on the parabola y = x ^ 2, then Y0 = x0 ^ 2
The distance from point P to line 2x-y-4 = 0:
d＝|2x0－y0－4|/√5＝|2x0－x0^2－4|/√5＝|－(x0－1)^2－3|/√5
Then when x0 = 1, D has a minimum value, Dmin = 3 / √ 5 = 3 √ 5 / 5
The shortest distance from the point of parabola y = x ^ 2 to the line 2x-y-4 = 0 is 3 √ 5 / 5

Find the coordinates of the shortest point on the parabola y & # 178; = 2x to the straight line X-Y + 3 = 0

Let the point on the parabola be p (y ^ 2 / 2, y). Then the distance s from P to the straight line X-Y + 3 = 0 = | y ^ 2 / 2-y + 3 | / √ 2 = | y ^ 2-2y + 6 | / (2 √ 2) = | (Y-1) ^ 2 + 5 | / (2 √ 2) > = 5 / (2 √ 2) when y = 1, x = 1 ^ 2 / 2 = 1 / 2, the distance s has the minimum value of 5 √ 2 / 4. At this time, the distance of the straight line of point P (1 / 2,1) is the minimum

The coordinates of the nearest point from the parabola y = X2 to the straight line 2x-y = 4 are ()
A. （32，54）B. （1，1）C. （32，94）D. （2，4）

Let P (x, y) be any point of parabola y = X2, then the distance from P to the straight line d = | 2x − y − 4 | 5 = | x2 − 2x + 4 | 5 = (x − 1) 2 + 35, | x = 1, D takes the minimum value 355, then p (1,1)