Given / M / = 5, / N / = 2, / M-N / = N-M, then the value of M + n is () A-7 B-3 Given / M / = 5, / N / = 2, / M-N / = N-M, then the value of M + n is () A-7 B-3 C-7 or - 3 D7 or - 7 or 3 or - 3

Given / M / = 5, / N / = 2, / M-N / = N-M, then the value of M + n is () A-7 B-3 Given / M / = 5, / N / = 2, / M-N / = N-M, then the value of M + n is () A-7 B-3 C-7 or - 3 D7 or - 7 or 3 or - 3

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Make two tangent lines of parabola x ^ 2 = 2PY (P > 0) through point m (2,2p). The tangent points are a and B respectively. If the ordinate of the midpoint of line AB is 6, then the parabola equation?
X = PK in the answer

A: the point m in your question should be (2, - 2P), so that you can make two tangents
Let the tangent equation passing through point m (2, - 2P) be y - (- 2P) = K (X-2), that is, y = kx-2k-2p
Substituting into the parabolic equation x ^ 2 = 2PY, we get: x ^ 2 = 2p (kx-2k-2p) = 2pkx-4pk-4p ^ 2
x^2-2pkx+4pk+4p^2=0…… （1）
A straight line is tangent to a parabola, which means that there is only one intersection point. The above formula has only one real number
△ = (- 2pk) ^ 2-4 (4pk + 4P ^ 2) = 0, that is: PK ^ 2-4k-4p = 0 （2）
According to Veda's theorem: K1 + K2 = 4 / P （3）
The unique solution of equation (1) x = - (- 2pk) / (2 * 1) = PK
Substituting into the parabolic equation x ^ 2 = 2PY, y = (PK) ^ 2 / (2P) = PK ^ 2 / 2 （4）
From (2) and (4), y = (4K + 4P) / 2 = 2 (K + P) （5）
Let point a (x1, Y1), point B (X2, Y2), and the ordinate of the midpoint of AB be: Y1 / 2 + Y2 / 2
So: 6 = Y1 / 2 + Y2 / 2 = (K1 + P) + (K2 + P) = K1 + K2 + 2p （6）
From (3) and (6), it is concluded that:
4/p+2p=6
The solution is p = 1 or P = 2
So the parabolic equation is: x ^ 2 = 2Y or x ^ 2 = 4Y

It is known that the focal point of the parabola y = x is f, the collimator is l, the tangent of the parabola is made through a point P on L, and the tangent points are a and B respectively?
In the process of solving the problem, one step is that the equation of PA is the square of y-x1 = 2x1 (x-x1). How do you deduce this step?

Y = x & # 178; derivation y '= 2x let a (x1, X & # 178; 1), B (X2, X & # 178; 2) x1 ≠ x2  curve at a tangent slope Ka = y' | (x = x1) = 2x1 〉 PA equation is Y-X & # 178; 1 = 2x1 (x-x1) similarly, Pb equation is Y-X & # 178; 2 = 2x2 (x-x2) P on the Quasilinear, let (m, - 1 / 4) ≠ - 1 / 4-x & # 178; 1 = 2x

Making a parabola through point m (a, 1) and finding two tangent lines Ma MB A. B of x 2 = 4Y as tangent points to prove that A. B passes through a fixed point and coordinates

The equation of straight line passing through M is y = K (x-a) + 1
Take the parabolic equation, Δ = 0, and get the relationship between K and a, as well as the coordinates of a and B (expressed by a)
The AB equation is y = (A / 2) X-1
So AB passes the fixed point (0, - 1)

High middle parabola: given points a (0, - 3), B (2,3), point P on x = y, find the minimum value of triangle PAB

See AB as the bottom
It only needs to solve the problem of height minimum
When the line parallel to AB is tangent to the parabola, the height is the minimum
l ab
y=3x-3
k=3
Parabola y = x ^ 2
Derivation y = 2x = 3
x=1.5
So p (1.5,2.25)
Tangent y = 3x-2.25
Distance between two straight lines d = 3 / (4 root sign 10)
So s = 1 / 2 * radical 40 * 3 / (4 radical 10) = 0.75

Given that a (0, - 1), B (3,2), P is any point of parabola y = 3x ^ 2 + 1, find the minimum area of △ PAB and the coordinates of point P at this time

AB = 3 √ 2; ab line y = X-1 = > x-y-1 = 0; let the height of P (T, 3T ^ 2 + 1) AB = distance from P to ab = | t-3t ^ 2-1 | / √ 2 = | 3T ^ 2-T + 2 | / √ 23t ^ 2-T + 2 = 3 (T - 1 / 6) ^ 2 + 23 / 12 > = 23 / 12 = > minimum height = (23 / 12) / √ 2 = (23 √ 2) / 24 minimum area = (1 / 2) * (3 √ 2) *

If the line y = - 2 and the parabola y = - X & sup2; intersect at two points a and B, and the point P is on the parabola y = - X & sup2;, if the area of △ PAB is two roots 2, the coordinates of P are obtained

The intersection points are: (√ 2, - 2), (- √ 2, - 2)
｜AB｜＝√2－（－√2）＝2√2
Let P (x, - x ^ 2)
S△PAB＝｜AB｜*|-X^2+2|/2
（2√2）*|-X^2+2|/2＝2√2
|2-x^2|=2
When 2 ≥ x ^ 2, 2-x ^ 2 = 2, x = 0, y = 0
two

It is known that P and Q are two points on the parabola x2 = 2Y. The abscissa of points P and Q are 4, - 2 respectively. If the two tangents cross point a, the ordinate of point a is ()
A. 1B. 3C. -4D. -8

∵ P and Q are two points on the parabola x2 = 2Y, the abscissa of points P and Q are 4, - 2, ∵ P (4,8), q (- 2,2), ∵ x2 = 2Y, ∵ y = 12x2, ∵ y ′ = x, ∵ the slope of tangent equation AP and AQ is KAP = 4, KAQ = - 2, ∵ the tangent equation AP is Y-8 = 4 (x-4), that is y = 4x-8, the slope of tangent equation AQ is Y-2 =

Given that the distance from a point P to the collimator on the parabola X & # 178; = - 2Y is 3, then the coordinate of point P is a process

If the distance from a point P on the parabola X & # 178; = - 2Y to the collimator is 3, then the coordinate of point P is
The Quasilinear of parabola X & # 178; = - 2Y is y = 1 / 2
Let P (x, y), the distance from P to the guide line be 1 / 2-y
1 / 2-y = 3
∴y=-5/2
∴x²=-2*(-5/2)=5
∴x=±√5
Then p (± √ 5, - 5 / 2)

It is known that the abscissa of the two points P and Q on the parabola y = 1 / 2x & # 178 are 4, - 2 respectively, and the tangents passing through P and Q are the tangent lines of the parabola respectively. If the two tangents intersect at point a, then the ordinate of point a

Because the abscissa of points P and Q are 4, - 2, respectively,
The ordinates of P and Q are 8 and 2, respectively
Y = 1 / 2 * X2, so y ′ = x,
The slopes of the tangent of the parabola passing through points P and Q are 4, - 2, respectively,
So the tangent equations of parabola passing through point P and Q are y = 4x-8 and y = - 2x-2 respectively
The solution of simultaneous equations is x = 1, y = - 4
So the ordinate of point a is - 4