The parabola C1: y = 18 (x + 1) 2-2 is rotated 180 ゜ around the point P (T, 2) to obtain the parabola C2. If the vertex of the parabola C1 is on the parabola C2, and the vertex of the parabola C2 is on the parabola C1, the analytical expression of the parabola C2 is obtained

The parabola C1: y = 18 (x + 1) 2-2 is rotated 180 ゜ around the point P (T, 2) to obtain the parabola C2. If the vertex of the parabola C1 is on the parabola C2, and the vertex of the parabola C2 is on the parabola C1, the analytical expression of the parabola C2 is obtained

∵ y = 18 (x + 1) 2-2 vertex coordinates are (- 1, - 2), ∵ rotate 180 ゜ around the point P (T, 2), the vertex coordinates of parabola C2 are (2t + 1,6), ∵ the analytic formula of parabola C2 is y = - 18 (x-2t-1) 2 + 6, ∵ the vertex of parabola C1 is on the parabola C2, ∵ - 18 (- 1-2t-1) 2 + 6 = - 2, the solution is T1 = 3, T2 = - 5, ∵ the analytic formula of parabola C2 is y = - 18 (X-7) 2 + 6 or y = - 18 (x + 9) 2 + 6 .
Given the equation y = MX + m ^ 2 of the straight line L, the vertex of the parabola C1 and the center of the ellipse C2 are at the origin of the coordinate, and their focal points are on the y-axis,
When m = 1, there is only one common point between the straight line L and the parabola C1
Let the parabolic equation be X & # 178; = 2px (because the focus of the parabola is on the Y axis and the vertex is at the origin)
When m = 1, the linear equation is y = x + 1, that is, x = Y-1
Y & # 178; - 2Y (1-p) + 1 = 0, because there is only one common point between the straight line L and the parabola C1
Δ = 0, i.e. B & # 178; - 4ac = 4 (1-p) &# 178; - 4 = 0. I.e. 8p & # 178; - 8p = 0 (because P ≠ 0), then p = 1
Then the parabolic equation is: X & # 178; = 2Y
It is known that hyperbola C1: x ^ 2 / A ^ 2-y ^ 2 / 2A ^ 2 = 1 (a > 1), the vertex of parabola C2 is at the origin o, and the focus of C2 is the right focus of C1
You want C2, right?
According to C1, the right focus is (√ 3a, 0)
C2 focus is (P / 2,0)
Then p / 2 = √ 3a
p＝2√3a,
c2：y^2＝2px
Namely:
y^2＝4√3ax
We know that x > 0, and prove that x > SiNx
Let f (x) = x-sin x, and then derive f '(x) = 1-cos x, because x > 0, so f' (x) is greater than or equal to 0, so f (x) is an increasing function! F (x) > F (0) = 0, so x > SiN x
【sin(α+2kπ）+cos（π/2+α）+tan(3π-α）】/【sin(α-π）+cos（α-π/2)+cot(π/2-α）】
【sin(α+2kπ）+cos（π/2+α）+tan(3π-α）】/【sin(α-π）+cos（α-π/2)+cot(π/2-α）】
=(sina-sina-tana)/(-sina+sina+tana)=-1
Known: x > 0, prove x > SiNx
f(x)=x-sinx
f'(x)=1-cosx
Obviously, 1-cosx > 0
So it's an increasing function
So x > 0
f(x)>f(0)+0
So x > SiNx
Let f (x) = x-sinx
Derivation: F '(x) = 1-cosx
Because of cosx
If we know that Tana and tanb are two of the quadratic equations ax & # 178; + BX + C = 0 (B ≠ 0), then cot (a + b)=
tan(a+b) = (tana+tanb)/(1-tana*tanb)
cot(a+b) = (1-tanatanb)/(tana+tanb)
By Weida theorem
tana+tanb = -b/a
tana*tanb = c/a
Bring it into the solution
How to get x-sinx ~ x ^ 3 / 6 with lobita's law?
Ask for detailed explanation~
Let's look at the limit of (x-sin) / x ^ 3
Because the numerator denominator tends to zero, we use the law of lobita to obtain the upper and lower derivatives
(1-cosx)/3x²
Up and down still tend to 0, use again
sinx/6x
Finally, it can be obtained by one more up and down derivation
1/6
So LIM (x-sinx) / X & # 179; = 1 / 6
That is x-sinx ~ x & # 179 / 6
From the meaning of the title, we can get the conclusion
lim(x->0)(x-sinx)/(x^3/6)
=lim(x->0)(1-cosx)/(x^2/2)
=lim(x->0)(sinx)/(x)
=lim(x->0)(cosx)/1
=cos0
=1
Known 0
Δ = k ^ 2-4 (K + 1) ≥ 0k ^ 2-4k-4 ≥ 0k ≥ 2 + 2 √ 2 or K ≤ 2-2 √ 21) Sina + cosa = ksina * cosa = K + 1 (Sina) ^ 2 + (COSA) ^ 2 = (Sina + COSA) ^ 2-2sinacosa = 1K ^ 2-2 (K + 1) = 1K ^ 2-2k-3 = 0 (K-3) (K + 1) = 0k = 3 or K = - 1, 2) from 1), 2), k = - 1y = x ^ 2 + kx-k / 4 = x ^ 2-x +
It is known that the function f (x) is a decreasing function defined on R, and for any real number x, y satisfies f (x + y) = f (x) + F (y), f (1) = 1. If f (x) satisfies the inequality f (2x + 1) > F (x) + 2, then the value range of real number x is
f(x+y)=f(x)+f(y),f(1)=1
Then f (x + 1) = f (x) + F (1) = f (x) + 1
f(x+2)=f(x+1+1)=f(x+1)+f(1)=f(x)+1+1=f(x)+2
So f (2x + 1) > F (x) + 2 = f (x + 2)
Since the function f (x) is a decreasing function defined on R, it has
2x+1