Given that the parabola y = x2 + 2 (K + 1) x-k has two intersections with the X axis, and the two intersections are on both sides of the line x = 1, then the value range of K is______ ．

Given that the parabola y = x2 + 2 (K + 1) x-k has two intersections with the X axis, and the two intersections are on both sides of the line x = 1, then the value range of K is______ ．

∵ parabola y = x2 + 2 (K + 1) x-k has two intersections with X axis, and the two intersections are on both sides of the straight line x = 1, ∵ when x = 1, y ＜ 0, so substituting x = 1 into the analytical formula, we get: 1 + 2 (k + 1) - K ＜ 0 ∵ K + 3 ＜ 0, we get k ＜ - 3; so the value range of K is k ＜ - 3

According to the following conditions, write the parabolic standard equation (1) focus f (3,0) (2) quasilinear equation x = 1 / 4 (3) the distance between focus and quasilinear is 2, and the focus is on the X axis

1、
p/2=3
2p=12
x²=12y
2、
x=p/2=1/4
2p=1
y²=-x
3、
p=2
2p=4
So x & # 178; = - 4Y or x = 4Y

Write the standard equation of parabola according to the following conditions (1) the focus is f (3,0)

(1) the focus is f (3,0) y ^ 2 = 12x

If the distance between a and the directrix of the parabola is 4, then the length of AB is?
The answer is 16 / 3

The focus coordinate of parabola C is f (1.0)
The Quasilinear equation is x = - 1
Let the equation of line I be y = KX + B
Substituting f into y = kx-k
From the properties of parabola, we know that the distance from a to the Quasilinear is equal to the distance from a to the focus, that is, iafi = 4, and the abscissa X1 of point a = 4-1 = 3
Let a (3, Y1) B (X2, Y2)
If AF & # 178; = Y1 & # 178; + (3-1) & # 178; = 16, Y1 = 2 √ 3 (negative value rounding off)
Substituting the coordinates of point a into the linear equation, the linear equation k = √ 3, y = √ 3 (x-1)
Linear and Parabolic Equations 3 (x-1) &# 178; = 4x 3x & # 178; - 10x + 3 = 0, the sum of abscissa of point a and B is 10 / 3
From the properties of parabola, we know that the distance from a (b) to the Quasilinear is equal to the distance from a (b) to the focus, so iabi = iafi + ibfi = 3 + x2 = 10 / 3

If the line L passing through the focus of the parabola y ^ 2 = 4x intersects the parabola at two points a and B, | ab | = 8, then the inclination angle of the line L is

Do it in polar coordinates
Taking the focus of the parabola as the polar origin,
ρ1=2/(1-cosθ) (1)
ρ1=2/(1-cos(θ+π)) (2)
ρ1+ρ2=8 (3)
Take (1), (2) into (3) to get the solution of θ
That's what we want

Higher numbers: how to solve the system of differential equations {DX / dt = x + 7Y? {dy / dt = 4x-2y

I'm sorry to miss the following equation just now. DX / dt = x + 7Y, ① dy / dt = 4x-2y, ② △ ① we get dy / DX = (4x-2y) / (x + 7Y) = [4-2 (Y / x)] / [1 + 7 (Y / x)] ③ equation ③ this is a homogeneous equation. The solution is as follows: let u = Y / x, then y = UX, dy / DX = u + X * Du / DX, so equation ③ becomes: U + X * Du / DX = (4-2u

Find integral DX / A & # 178; + X & # 178;

∫［1/（a^2＋x^2）］dx
＝（1/a^2）∫｛1/［1＋（x/a）^2］｝dx
＝（1/a）∫｛1/［1＋（x/a）^2］｝d（x/a）
＝（1/a）arctan（x/a）＋C

Given that f (a) = 0, f is continuously differentiable in the closed interval a-b. it is proved that ∫ (a to b) F & # 178; (x) DX ＜ = (B-A) &# 178; (2 ∫ (a to b) (f '(x)) &# 178; DX

For the convenience of narration, a = 0
That's right

If in the interval [- A, a], the definite integral ∫ [f (x)] &# 178; DX = 0, then is there always f (x) = 0?

If f (x) is a continuous function, then f (x) = 0
But if f (x) is only integrable but discontinuous, then f (x) ≠ 0

F (x) = 1 / X & # 178; + 1 + X & # 179; ∫ (0-1) f (x) DX find ∫ (0-1) f (x) DX

Let a = ∫ (0-1) f (x) DX
If you write 1 / (x ^ 2 + 1), then f (x) = 1 / (x ^ 2 + 1) + ax ^ 3
∫(0-1)f(x)dx=arctanx+ax^4/4(0-1)=a/4+π/4=a
So a = π / 3
If you write 1 / x ^ 2 + 1, then the integral of 1 / x ^ 2 on (0,1) does not converge and has no solution