It is proved that the coordinates of the point P which makes the vector AP * vector BP get the minimum value is [emergency] given that the point a (2,0), B (4,0), and the moving point P move in the parabola y ^ 2 = - 4x, it is proved that the coordinate of the point P which makes the vector AP * vector BP get the minimum value is the coordinate origin. Please give the detailed process,

It is proved that the coordinates of the point P which makes the vector AP * vector BP get the minimum value is [emergency] given that the point a (2,0), B (4,0), and the moving point P move in the parabola y ^ 2 = - 4x, it is proved that the coordinate of the point P which makes the vector AP * vector BP get the minimum value is the coordinate origin. Please give the detailed process,

Let p be (- y ^ 2 / 4, y) (y = 0
Vector AP * vector BP = (M + 2) (M + 4) + 4m = m ^ 2 + 10m + 8
When m = 0, the minimum value is obtained
At this point, x = 0, y = 0

Given the point a (- 2,0), B (2,0), the moving point P on curve C satisfies vector AP times vector BP = - 3 (1) The curve of equation C (2) If the straight line L passing through the fixed point m (0, - 2) has an intersection point with curve C, the value range of the slope of the straight line l can be calculated （3） If the moving point Q (x, y) is on the curve C, find the value range of U=y+2/x

1> Given the point a (- 2,0), B (2,0) let point P = (x, y). Therefore, vector AP = (x + 2, y), vector BP = (X-2, y). Therefore, the vector AP multiplied by the vector BP = x? 2 + y? - 4 = - 3, that is, x? + y? = 1. Therefore, the equation of curve C is: x? + y? = 1, and M = (0, - 2)

The point P of a (2,0) B (4,0) moves on the parabola y? = - 4x so that the vector AP is multiplied by the vector BP to get the P coordinate of the minimum value

Let P (x, y), X ≤ 0 and y? = - 4xap = (X-2, y), BP = (x-4, y) AP.BP= (X-2) (x-4) + y? = x? - 6x + 8-4x = x? - 10x + 8 = (X-5) Ω - 17x ≤ 0, y = (X-5) Ω - 17 is a decreasing function on (- ∞, 0), so when x = 0, AP.BP There is a minimum value of 8, where p (0,0)

We know that the vector OA = (2,2), vector ob = (- 4,1) point P on the non negative semiaxis of the X axis, O is the origin. 1. When the vector PA * Pb gets the minimum value Let the angle APB = θ, when point P satisfies 1, find the value of cos θ

If P ≥ 0, then the vector PA = op-oa = (P-2, - 2) and the vector Pb = op-ob = (P + 4, - 1) 1. The scalar product PA * Pb = (P-2) (P + 4) + - 2 * - 1) = P 2 + 2p-6 = (P + 1) 2 when p = 0, the scalar product PA * Pb gets the minimum value - 6, and the vector OP = (0,0) 2

Known vector OA=（1，3）， OB = (3, - 1), and AP＝ 2 Pb, then the coordinates of point P are () A. （2，-4） B. （2 3，-4 3） C. （7 3，-1 3） D. （-2，4）

Let the coordinates of point p be (x, y), and
AP＝ 2
The results show that - 3, X-1, X-1, X-1, X-1, X-1, X-1, X-1, X-1, X-1, X-1, X-1, X-1, X-1, X-1, X-1, X-1, X-1, X,
So X-1 = 6-2x, and Y-3 = - 2-2y, x = 7
3，y=-1
3, so the coordinate of point P is (7
3，-1
3），
Therefore, C

Let OA = (2,3), OB = (6, - 3) point P on the extension line of line Ba, and | AP | = 2 / 3 | Pb | to find the coordinates of P

Let P (x, y)
Then (x + 2) ^ + (3 + y) ^ = 2 / 3 [(x + 6) ^ + (Y-3) ^]
The solution is 1/3x^-4x-1/3y^+10y-17=0 ①
And because: X / y = (2-6) / [3 - (- 3)] = - 2 / 3 2
The value of XY can be obtained by solving formula 1 and 2

The straight line L passing through the point P (- 1, - 6) and the parabola y ^ 2 = 4x intersect two points a and B, if | AP | = | BP |

Let a (x1, Y1) and B (X2, Y2), then the coordinates of a and B satisfy parabola, and then: (Y1) 2 = 4 (x1) (Y2) 2 = 4 (x2) are subtracted from each other to obtain: (Y1 + Y2) (y1-y2) = 4 (x1-x2) (y1-y2) / (x1-x2) = 4 / (Y1 + Y2) consider: the slope of line AB is k =

The straight line PCD passes through point O, the chord AB is perpendicular to the diameter CD on e, and the point F is on the circle O, ∠ FPO = ∠ ofE. It is proved that PA is the tangent of circle o Draw your own picture

In △ OFP and △ OEF, ∠ FPO = ∠ EFO, ∠ POF = ∠ peo,
So, △ OFP ≌ △ OEF,
It can be concluded that: of: OE = OP: of;
Moreover, OA = of,
OA ∶ OE = op ∶ OA
In △ OAE, △ OAE = OAE,
So, △ OPA ≌ △ OAE,
It can be concluded that: ∠ OAP = ∠ OEA = 90 °,
So PA is tangent to circle o

As shown in the figure, points c and D are on line AB, and △ PCD is an equilateral triangle (1) When AC, CD and DB satisfy the relationship, ACP ∽ PDB; (2) When △ ACP ∽ △ PDB, calculate the degree of ∠ APB

(1) When CD2 = AC · dB, △ ACP ∷ PDB,
∵ △ PCD is an equilateral triangle,
∴∠PCD=∠PDC=60°，
∴∠ACP=∠PDB=120°，
If CD2 = AC · dB, PC = PD = CD: PC · PD = AC · dB,
PC
BD=AC
PD，
Then △ ACP ∷ PDB is obtained according to the judgment theorem of similar triangles
(2) When △ ACP ∽ PDB, ﹤ APC = ∠ PBD
∵∠PDB=120°
∴∠DPB+∠DBP=60°
∴∠APC+∠BPD=60°
∴∠APB=∠CPD+∠APC+∠BPD=120°
The degree of ∠ APB is 120

PAB and PCD are secants of circle O, PA = Pb, and ab = CD Don't use the secant theorem. We didn't learn it. It's wrong on the Internet! All right, great reward! Urgent

It should be pa = PC
prove:
Make OE ⊥ PAB in E
Do of ⊥ PCD in F
PA=PC, OP=OP, OA=OC ==> △POA≌△POC
∠OPA=∠OPC
That is, OP is the angular bisector of APC
Then OE = of
Congruence (HL) of two right triangles with equal hypotenuse and constant angle
Then △ EOA ≌ △ FOC = = > AE = CF
△EOB≌△FOD ==> EB=FD
AE+EB=AB
CF+FD=CD
So AB = CD
The proof is over