If F1 and F2 are the left and right focal points of the hyperbola, P is a point on the Quasilinear, Pf1 vector ⊥ PF2, Pf1 × PF2 = 4ac, then E is equal to

F1 and F2 are the left and right focuses of hyperbola, and P is a point on the right collimator. If P (a ^ 2 / C, m) ∵ Pf1 vector ⊥ PF2, | Po | = √ [a ^ 4 / C ^ 2 + m ^ 2] = C ∧ a ^ 4 / C ^ 2 + m ^ 2 = C ^ 2, then m ^ 2 = C ^ 2-A ^ 4 / C ^ 2 ∧ Pf1 × PF2 = 4ac | Pf1 | ^ 2 | PF2 = 16A ^ 2C ^ 2 ∧ [(c + A ^ 2 / C) ^ 2 +

If there are two different intersections between the line L: y = x + B and the curve y = √ 4-x & # 178, find the value range of the real number B

The curve y = √ 4-x & # 178; the square of both sides: y ^ 2 = 4-x ^ 2, (Y ≥ 0, - 2 ≤ x ≤ 2)
Sort out x ^ 2 + y ^ 2 = 4, (Y ≥ 0, - 2 ≤ x ≤ 2), (1) formula
It is a semicircle with y axis as symmetry axis and X axis as top, including two points on X axis
This is the problem of the intersection of a straight line and a semicircle
The slope of the straight line is 1. When B = 2, the straight line and the semicircle intersect at two points (0,2), (- 2,0). When the straight line continues to move upward (b > 2) and is tangent to the semicircle, then the intersection point is 1. At this time, y = x + B is substituted into (1)
X ^ 2 + (x + b) ^ 2 = 4, there is only one root, B = ± 2, 2 under root, negative value is rounded off, B = 2 √ 2
In conclusion, 2 ≤ B

When the curve y = √ (1-x & # 178;) and the straight line y = K (x-1) + 1 have two common points, find the value range of the real number K
A. (0, positive infinity)
B.（0,1）
C.[0,1/2]
D.(0,1/2]
I know how to choose D

It's better to draw a picture first and then do it,
Obviously, the curve y = √ (1-x & # 178;) represents a semicircle with radius 1 and center (0,0),
The line y = K (x-1) + 1 must pass through (1,1) point
Obviously, only when the line is connected between (1,1) and (0,1) and (1,1) and (- 1,0),
Only curves and lines have two common points,
The slope of the line connecting (1,1) and (0,1) is 0,
The slope of the line between (1,1) and (- 1,0) is 1 / 2
When the slope is 0, there is only one intersection between the curve and the line, and when the slope is 1 / 2, there are two intersections between the curve and the line
So the slope k is in the range of (0,1 / 2)], select D

Knowing the joint density function of X and y, how to find the probability density function of Z = x + y
Z = x + YX ~ u [0,1] y ~ u [- 1,0] (U means uniform distribution) to find the density function of Z? It should use convolution formula, but I don't know what the upper and lower limits of integration should be?
I know that the probability density function f (x) of X is 1, the probability density function f (y) of Y is 1, and the joint probability density f (x, y) = f (x) f (y) of X and Y is also 1. Therefore, the distribution function f (z) of Z is f (x, y) DXDY, where the integral region is the lower left part of the square (0 ≤ x ≤ 1; - 1 ≤ y ≤ 0) in X + y = Z. so, how to find f (z)? Don't give the answer directly!

You can use their two ranges to express the relationship between X and Z, that is to say, draw a region in the coordinate system with Z as the horizontal axis and X as the vertical axis. Finally, you can use ∫ f (x, z-x) DX to integrate X. the upper and lower lines are the ranges of X, and Z is used to express them. This is the result, but you should pay attention to the value range of Z. in addition, convolution formula can't be used casually, and it's easy to make mistakes, So let's just use the original definition

The range of y = Sin & # 178; x-3cosx + 1 / 4

y=1-cos²x-3cosx+1/4
=-(cosx+3/2)²+7/2
-1

Is there a function that is both odd and even?
Is there a function that is both odd and even (both odd and even functions)

Y = 0, the domain is symmetric with respect to the coordinate origin
There are countless such functions
As long as y = 0, the domain of the independent variable x is symmetric about the Y axis
It can be thought that if a point on the function is not on the coordinate axis, that is, in the quadrant, then it is symmetrical about the origin and about the Y axis, so the two points share one abscissa and are not functions
Only one origin can be obtained on the y-axis, and symmetry points are required at other positions
To sum up, it can only be taken on the x-axis

How to choose an arc of ellipse in Geometer's Sketchpad

In the latest version of 5.03 strongest Chinese version, which was updated on August 31, there is a tool to draw an arc of ellipse in "custom tool" - conic curve C "

As shown in the figure, extend the edge ab of rectangle ABCD to point E, so that AE = AC, f is the midpoint of CE,

Man, what about your picture? Ah, it's a question and a picture. What's the situation

prove:
Connect BD & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;
∵ AE = AC, AC = DB
∴ AE=DB
∵ △ BCE is a right triangle, and point F is the midpoint of CE
So BF = CF = EF & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; BCF = CBF
And ∠ ACB = ∠ DBC & nbsp; & nbsp; & nbsp; & nbsp; DBF = ∠ ace
AC = AE & nbsp;, ∠ AEC = ace
∴ ∠DBF=∠AEC
From the edge, corner and edge theorem, △ AEF ≌ △ DBF is deduced
In △ ace, AC = AE, point F is the midpoint of CE, and ⊥ AF ⊥ CE, i.e., ∠ AFE = 90 degree
It is proved that DF ⊥ BF = 90 degree

How to find (c) in AX + BX + C of quadratic function
Make it simple
Is the constant term of a quadratic function

When x = 0, y = C
C is the ordinate of the intersection of the parabola and the y-axis

There are three piles of pieces, each pile has the same number of pieces, and only black and white pieces
There are three piles of chessmen, each pile has the same number of chessmen, and there are only black and white chessmen. The first pile has the same number of black and white chessmen as the second pile, and the third pile accounts for 2 / 5 of all the black ones?

Let all the sunspots be x and all the white ones be y, then each pile of chess pieces is (x + y) / 3; because there are as many sunspots in the first pile as in the second pile, and each pile of chess pieces is as many, the first pile and the second pile share (x + y) / 3 sunspots
(x + y) / 3 + 2x / 5 = x, x = 5Y / 4, Y / (x + y) = Y / [(5Y / 4) + y] = 4 / 9

If F1 and F2 are the left and right focal points of the hyperbola, P is a point on the Quasilinear, Pf1 vector ⊥ PF2, Pf1 × PF2 = 4ac, then E is equal to