From a point P on the parabola y2 = 4x, the vertical foot of the parabola is m, and | PM | = 5. Let the focus of the parabola be f, then the area of △ MPF is______ ．

From a point P on the parabola y2 = 4x, the vertical foot of the parabola is m, and | PM | = 5. Let the focus of the parabola be f, then the area of △ MPF is______ ．

Let P (x0, Y0) according to the meaning of the problem, we can see that the parabola quasilinear x = - 1, ∧ x0 = 5-1 = 4, ∧ Y0 | = 4 × 4 = 4, and the area of ∧ MPF is 12 × 5 × 4 = 10, so the answer is 10

High school mathematics elective course - parabola through parabola y & # 178; = 2px focus F and slope of four-thirds of the straight line intersection
If the parabola passing through the focus F of the parabola y & # 178; = 2px with a slope of four-thirds intersects with two points a and B, if the vector AF = λ vector FB (λ > 0), then the value of λ is
A、5
B、4
C、4/3
D、5/2
Please tell me which one to choose and why,

Choose B
y/（x-p/2）=4/3
y^2=2px
The simultaneous results are x = 2p and x = 1 / 8p
Add a P / 2 to it all
The idea is to transform the multiple of vector into the multiple of length, and then transform it into the distance between a and B points and the guide line
Because it is a multiple choice question, the more concise way is to set p to 2, the calculation is easier, the result is the same

A mathematical problem. Given that the point m (3,2), f is the focus of the parabola y & # 178; = 2x, the point P moves on the parabola. When the value of | PM | + | PF | is the minimum, the coordinates of P

(2,2) [analysis] if the perpendicularity is D, then | PF | = | PD | PM | + | PF | = | PM | + | PD | ≥ | MD | is the same sign if and only if the three points m, P and D are collinear. Therefore, when p is the intersection of the perpendicularity and the parabola, the minimum value can be obtained. At this time, P (2,2) your praise is my progress

If the distance from a point m on the parabola y = 4x2 to the focus is 1, then the ordinate of point m is ()
A. 1716B. 1516C. 78D. 0

According to the definition of parabola, if the distance from m to the focus is 1, then the distance from m to the collimator is also 1. The collimator of ∵ parabola is y = - 116, and the ordinate of ∵ m point is 1-116 = 1516

Parabola y = 4x ^ 2. The distance from the previous point m to the focus is 1. What is the distance from the m point to the Y axis?
17 / 16 of item a. Item B 1. The root of item C is 15. 15 / 16 of item D.

Parabola X & # 178; = (1 / 4) y
Focus f (0,1 / 16)
Quasilinear equation y = - 1 / 16
Set point m (x, y)
According to the proposition and the definition of parabola
y+(1/16)=1
∴y=15/16
The distance from point m to the y-axis is 15 / 16
Choose D

If the distance from a point P on the parabola y ^ 2 = - 4x to the focus is 4, then its abscissa is

According to the definition of parabola:
The distance from P to the guide line is 4
∵y²=-4x
The guide line: x = 1
The abscissa of P is - 3

The distance from a point on the parabola y ^ 2 = - 4x to the focus is 4. Find its abscissa

Distance to focus = distance to guide line
y^2=-4x
Focus (- 1,0) collimator x = 1
Abscissa x
1-x=4
x=-3

The movement of the two ends of the line segment AB of fixed length 3 on the parabola y ^ 2 = x, the midpoint of the line segment AB is m, the shortest distance from the point m to the Y-axis and the coordinates of M are calculated

Let a and B coordinates be (x1, Y1), (X2, Y2) respectively, then the distance from m point to y axis is f (x1, x2) = (x1 + x2) / 2. According to the subject conditions, we can get the following equation: Y1 ^ 2 = x1y2 ^ 2 = X2 (x1-x2) ^ 2 + (y1-y2) ^ 2 = 3 ^ 2, and get: X1 ^ 2-2x1x2 + x2 ^ 2 + X1 + x2-2 √ (x1x2) = 9, let φ (x1, x2) = X1 ^ 2-2x1x2 + x2 ^ 2

The two ends of the line segment AB with fixed length of 3 move on the parabola y ^ 2 = x, mark the midpoint of the line segment AB as m, calculate the shortest distance from the point m to the Y axis, and calculate the M coordinate at this time. The abscissa is five fourths. How to calculate the ordinate?

According to the midpoint ordinate formula
5 (Y1 + Y2) = ± root 2 / 2

Given that the length of line AB is 3, both ends are on the parabola x = y ^ 2, try to find the coordinates of m when the shortest distance from the midpoint m of AB to the Y axis

Method 1
Let a be (yo ^ 2, yo) and B be (Y1 ^ 2, Y1)
So the midpoint coordinates are ((yo ^ 2 + yo) / 2, (Y1 ^ 2 + Y1) / 2)
According to the distance formula (yo ^ 2-y1 ^ 2) ^ 2 + (y0-y1) ^ 2 = 9
According to the distance from the midpoint to the Y point, d ^ 2 = ((Y1 ^ 2 + Y0 ^ 2) / 2) ^ 2
Simplify the equation of D with respect to Y0 by substituting formula (1) into formula (2), and obtain the derivative so that the derivative is equal to zero to obtain the value of Y0
Then, y 0 is substituted into 1, so that the coordinates of M are (9 / 4,0)
Method 2
Because the parabola is symmetric about the X axis, the distance from the midpoint to the Y axis is the shortest when AB is parallel to the Y axis. The m point is on the X axis, so let a (yo ^ 2, Y0) B (Y0 ^ 2, - yo)
The root is based on the distance formula of two points: 2y0 = 3
So the coordinate of M is (9 / 4,0)