Let the coordinates of the two chords be y x 2, y x 1, y x 2 = 2|=______ ．

Let the coordinates of the two chords be y x 2, y x 1, y x 2 = 2|=______ ．

∵ parabola y2 = 4x | P = 2 according to the definition of parabola, we can get | ab | = X1 + x2 + P = 6 + 2 = 8, so the answer is: 8

The vertex of the parabola is at the origin, the focus is the left focus of the ellipse X & sup2 / 5 + Y & sup2; = 1, and the chord of the parabola is drawn through the point m (- 1, - 1), so that M is the midpoint of the chord, and the chord length of the chord is calculated
Can you give me a little process? There are still a few questions, please do it~
What is the structure of the last formula?

The focus of the ellipse is (- 2,0), so the equation of the parabola is y ^ 2 = - 8x. Obviously, the slope of the straight line exists, so let the equation be y + 1 = K (x + 1). Then, the equation of the straight line and the equation of the parabola are established, and Y is eliminated. A quadratic equation of one variable about X is obtained. By using WIDA's theorem, X1 + x2 = - 2, the value of K can be obtained

Application and practice of the second definition of conic ellipse in high school mathematics

The ratio of the distance from the plane to the fixed point F (C, O) and the fixed line x = a ^ 2 / C is constant. The locus of the point m of C / A is ellipse

X tends to infinity LIM (e ^ x-2x) / (e ^ x + 3x)
Isn't the 0 / 0 model able to use the lobita rule?

It's in the form of "infinity / Infinity" and it can also be done with Robida. You can do this by dividing the numerator and denominator by X to get ((e ^ X / x) - 2) / ((e ^ X / x) - 3)
And e ^ X / X is positive infinity when X - > infinity, which can be obtained by using Robida. Then the original formula is obviously infinite / Infinity form, which satisfies the condition, and can be used as l; hospital's
Using L'Hospital's rule to derive the numerator denominator twice, we get e ^ X / e ^ x = 1

X approaches to 0, K, N, m are all constants. Find the limit of LIM [(tankx) ^ n] / [(SiNx) ^ m]

KX and X tend to zero
Then tankx ~ KX
sinx~x
So the original formula = LIM (KX) ^ n / x ^ m
=limk^n*x^(n-m)
therefore
If nm, the original formula is 0

If a > 0, b > o are constants, then the limit LIM ((a ^ x + B ^ x) / 2) ^ (1 / x) of X tending to 0

This is the limit of type 1 ^ ∞. The important limit LIM (1 + x) ^ (1 / x) = elim [(a ^ x + B ^ x) / 2] ^ (1 / x) = Lim [1 + (a ^ x + B ^ X-2) / 2] ^ (1 / x) = Lim [1 + (a ^ x + B ^ X-2) / 2] ^ {[1 / (a ^ x + B ^ X-2)] * [(a ^ x + B ^ X-2) / x]} = e ^ Lim [(a ^ x + B ^ X-2) / x] = e ^ LIM (a ^ x * LNA + B ^ xlnb) = e ^ (

Find the limit: LIM (x → + ∞) √ x (√ (a + x) - √ x) (a ∈ R is a constant)

Multiplication √ (a + x) + √ x
The square difference is a + x-x = a
So the original formula = LIM (x → + ∞) a √ X / [√ (a + x) + √ x]
Divide up and down by √ x
=lim(x→+∞)a/[√(a/x+1)+1]
=a/(√a+1)

Given the line ax-y + 2A + 1 = 0, if Y > 0 is constant when x ∈ (- 1,1), the value range of a is obtained
It is known that ax-y + 2A + 1 = 0
(1) If Y > 0 holds when x ∈ (- 1,1), the value range of a is obtained
(2) If y ≥ 0 holds when a ∈ (- 1 / 6,1), the value range of X is obtained

1.
ax-y+2a+1=0
y=ax+2a+1
This is a linear equation, as long as the first and last endpoints are greater than 0
That is y (x = - 1) = a + 1 > 0
y(x=1) =3a+1>0
The solution is a > - 1 and a > - 1 / 3
It's a > - 1 / 3
2. If a ∈ (- 1 / 6,1), y ≥ 0 holds,
It's also a linear function, substituting the beginning and end points
y(a=-1/6)=-1/6x+2/3>=0
y(a=1) = x+9 >=0
Solution
x=-9
It is - 9

If the inequality x ^ 2 + 2 + | x ^ 3-2x | > = ax is (0,4) constant for X, find the value range of real number a
For x + 2 / x + | x ^ 2-2 | there is a minimum value when x = √ 2
Can we do without absolute value?

x^2+2+|x^3-2x|≥ax
Because x ∈ (0,4)
So divide both sides by X
So there are: x + 2 / x + | x ^ 2-2 | ≥ a
For x + 2 / x + | x ^ 2-2 | there is a minimum value when x = √ 2
So a ∈ (- ∞, 2 √ 2]
2 you can have one less condition (x > = 0) when x + 2 / X is the minimum, x = √ 2
|When x ^ 2-2 | = 0 is the minimum, when x = √ 2, because x = √ 2 is the minimum at the same time, it's OK not to go

Given that the domain of definition of function f (x) is r, and satisfies the following conditions: (1) when x > 0, f (x) 0, the inequality f (AX-2) + F (x-x ^ 2) > 0 holds, and the range of real number a is obtained

For any real number x and y, f (x + y) = f (x) + F (y)
So f (0) = f (0) + F (0). So f (0) = 0
f(0)=f(x+-x)=f(x)+f(-x)=0.
So f (- x) = - f (x) is an odd function
When x0, f (- x) 0
That is, when x > 0, f (x) 0 holds for x > 0
So AX-2 + x-x ^ 2