As shown in the figure, the curve C is the image of the function y = 6 / X in the first quadrant, and the parabola is the image of the function y = - x ^ 2-2x + 4, The point PN (x, y) (n = 1,2,3 ··) is on the curve C, and X and y are integers. (1) find out all the points PN (x, y); (2) take any two points in PN as a straight line, and find out the number of all different lines. (3) take any straight line from (2), and find out the probability that the straight line and parabola have a common point

As shown in the figure, the curve C is the image of the function y = 6 / X in the first quadrant, and the parabola is the image of the function y = - x ^ 2-2x + 4, The point PN (x, y) (n = 1,2,3 ··) is on the curve C, and X and y are integers. (1) find out all the points PN (x, y); (2) take any two points in PN as a straight line, and find out the number of all different lines. (3) take any straight line from (2), and find out the probability that the straight line and parabola have a common point

1) Because X and y are integers, X is a divisor of 6, that is, x = - 6, - 3, - 2, - 1,1,2,3,6,
Corresponding y = - 1, - 2, - 3, - 6,6,3,2,1,
So the points are P1 (- 6, - 1), P2 (- 3, - 2), P3 (- 2, - 3), P4 (- 1, - 6), P5 (1,6), P6 (2,3), P7 (3,2), P8 (6,1)
2) No three of the above eight points are collinear, so if any two points are taken as a straight line, 8 * 7 / 2 = 28 different straight lines can be made
3) Among the 28 straight lines mentioned above, there are 24 lines with common points with parabola and 4 lines without common points (p5p6, p5p7, p5p8, p6p7). Therefore, the calculated probability = 24 / 28 = 6 / 7

The parabola y = x2 + (M + 2) x-2m is known. When m = (), the parabola passes through the origin

The parabola goes through the origin, that is, the parabola goes through (0,0)
The point on the line substitutes (0,0) into the equation
So 0 = 0 + 0-2m, so m = 0

In the plane rectangular coordinate system, O is the coordinate origin, the line y = 3x + 4 intersects the Y axis at point a, whether there is a point P on the parabola y = 2x2, so that the area of △ POA is equal to 10? If it exists, find out the coordinates of point p; if not, explain the reason

Suppose that there is a point P (m, n), so that the area of △ POA is equal to 10; s = 12oa · | m |, that is, 10 = 12 × 4 × | m |, the solution is: | m | = 5, | M = 5 or - 5; substituting m into y = 2x2, the solution is n = 50, and the coordinates of point P are: (5, 50) or (- 5, 50)

The origin of the plane rectangular coordinate system is O, take a point P on the parabola y = 1 / 2x ^ 2, and take a point a on the X axis, so that OP = PA,
The origin of the plane rectangular coordinate system is O, take a point P on the parabola y = 1 / 2x ^ 2, take a point a on the X axis, so that OP = PA, make a vertical line of the X axis through point a and intersect with the straight line OP at Q. when △ Apq is an equilateral triangle, calculate the area of △ Apq

According to the title, if the angle OPA is 120 degrees and the angle AOP is 30 degrees, then let OP = a, the coordinates of point p be (+ - A / 2 times root 3, a / 2), and the area of △ Apq be a ^ 2 / 4 * root 3
The coordinates of point P are brought into the parabolic equation. A = 4 / 3, a not = 0 are obtained
The area of △ Apq is 4 / 9 times that of root 3

It is known that a and B are two points on the parabola y ^ 2 = 4x, O is the coordinate origin, OA is perpendicular to ob, and it is proved that the product of the ordinates of a and B is a constant

Let a (x1, Y1), B (X2, Y2), because OA is perpendicular to ob, a and B can not be in the same quadrant, if they are in the same quadrant, then the angle between OA and ob is less than 90 degrees, and they can only be in different quadrants one or four, so the ordinate symbols of a and B are opposite, vector OA = (x1, Y1), vector ob (X2, Y2), here let a be in the first quadrant

Is the odd function always symmetric about the origin? Is the even function always symmetric about the y-axis? Can it be proved

It can be proved that: 1. The odd function f (x) is characterized by the odd function property f (x) = - f (- x). Let a value of the independent variable be d (first, the domain of definition is symmetric about the origin) and f (d) = y, then the function f (x) passes (D, y) is characterized by the odd function property, and f (x) passes (- D, - y). These two points are related to the origin symmetry

The symmetry principle of odd function about origin and even function about y axis

Odd function f (x) = - f (- x)
Even function f (x) = f (- x)

∫ (x + y) DXDY, where D: X * x + y * y
I don't understand. I'm more detailed

r^2-2rcosa

A calculus exercise for freshmen and senior students proves that f (x) = 1 / x, cos (1 / x) is unbounded in (0,1)

Disprove proof
Suppose that for any e, there is always m in [0,1] such that [y-m]

The proof of the first year's Higher Mathematics
It is proved that when x → 0, there is arctanx ~ X

Let t = arctanx, then x = tank, X → 0, then t → 0, that is, when t → 0, t = tank, tank = sin / cost, tank / T = (sin / T) * (1 / cost), when t → 0, sin / T = 1,1 / cost = 1, so, tank / T = 1, it is proved. So, when t → 0, t = tank, that is, when x → 0, there is arctanx ~ X