Through the fixed point m (4,0), make a straight line L intersection parabola y ^ 2 = 4x at two points a and B, f is the focus of the parabola, and find the minimum value of △ Fab area?

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It is known that the focus of parabola C, y ^ 2 = 4x is F. the line L passing through f intersects with C at a and B. If AB equals 16 / 3, first, find the linear equation and second, find the minimum of ab

F (1,0), directrix: x = - 1, let L: y = K (x-1), bring in Y ^ 2 = 4x to get k ^ 2 * x ^ 2-2 (k ^ 2 + 2) x + K ^ 2 = 0. Two X 1 and x 2 are abscissa of two intersection points. From the definition of parabola, we know that ab = AF + BF = sum of distances from a and B to the directrix = X1 + x2 + 2, X1 + x2 = 16 / 3-2 = 10 / 3, and 2 (k ^ 2 + 2) = 10 / 3 *

It is known that the focus of the parabola y ^ = 2px (P > 0) is F. the line passing through f intersects Y axis at point P, and intersects the parabola at two points AB, where point a is in the first quadrant
(1) Verification: the circle with the diameter of line FA is tangent to the y-axis
(2) If the vector FA = λ 1, the vector AP, the vector BF = λ 2, the vector FA, λ 1 / λ 2 belong to the range of [1 / 4,1 / 2] to find λ 2

(1) Let the abscissa of point a be Xa, and D denote the distance from each point to the guide line, then Da = XA + P / 2, so AF = Da = XA + P / 2 = 2R, and the distance from the center of the circle to the y-axis do = (XA + XF) / 2 = (XA + P / 2) / 2 = R, so it is tangent

What is the distance from the focus of the parabola y ^ 2 = 20x to the collimator

y^2=20x=2px
So p = 10
Then the distance from the focus to the guide line = P = 10

The distance from the focus of the parabola y = 10x to the collimator is () a.2.5 B.5 c.7.5 d.10

It's B. the distance from the origin to the focus to the directrix is 2 / 2 P, so the distance from the focus to the directrix is p. the formula y = 2px 2p equals 10 P equals 5

If the distance from a point on the parabola to the focal point is 8, then the distance from this point to the directrix is 0,

The distance to the directrix is also 8, according to the definition of parabola
A parabola is one whose distance to the vertex is equal to that to the fixed line
A definite line is a directrix

Is the distance from a point of parabola to the focus equal to the distance from the collimator equal to half p?

Yes, that's right

What is the distance from the focus of the parabola y equal to 4x to the collimator

The distance from the focus (1,0) of the parabola y ^ 2 = 4x to the collimator L: x = - 1 is 2

Is it a theorem that the distance from any point on a parabola to the collimator is equal to the distance from it to the focus? What's its name?
What are the directrix and focus in the theorem?

This is not a theorem, but the definition of parabola. The set of points to a straight line and a point outside the line whose distance is equal is called a parabola. The straight line is called a collimator, and the point outside the line is the focus

The focal coordinate of parabola y = 8x2 is______ ．

The parabola y = 8x2 can be changed into x2 = 18y, the focus on the Y axis ∵ 2p = 18, ∵ 12p = 132 ∵ the focus coordinate of the parabola y = 8x2 is (0132), so the answer is: (0132)

Through the fixed point m (4,0), make a straight line L intersection parabola y ^ 2 = 4x at two points a and B, f is the focus of the parabola, and find the minimum value of △ Fab area?