From a point Q on the hyperbola x2-y2 = 1, the perpendicular of the straight line x + y = 2 is drawn. If the perpendicular foot is n, then the trajectory equation of the midpoint P of the line QN is______ ．

Let P (x, y), q (x1, Y1), then n (2x-x1, 2y-y1), ∵ n is on the straight line x + y = 2, ∵ 2x-x1 + 2y-y1 = 2, and ∵ PQ is perpendicular to the straight line x + y = 2, ∵ y − y1x − x1 = 1, that is, X-Y + y1-x1 = 0

From a point Q on the hyperbola x2-y2 = 1, the perpendicular of the straight line x + y = 2 is drawn. If the perpendicular foot is n, then the trajectory equation of the midpoint P of the line QN is______ ．

If n (2x-x1, 2y-y1) is n (2x-x1, 2y-y1), then n (2x-x1, 2y-y1) n (2x-x1, 2y-y1) is n (2x-x1, 2y-y1-y-1) if n (2x-x1, 2y-y1-y1) is on the straight line x + y = 2, and {2x-x1 + 2x-x1 + 2y-y1-y1 = 2 \\\\\\\\\\\\\\\\\\\\\\\\\\\\y2-2x + 2y-1 = 0 So the answer is: 2x2-2y2-2x + 2y-1 = 0

Given points a (1,2,1); B (- 1,3,4); C (1,1,1), if AP vector = 2 times Pb vector, then the module length of PC vector is?

Let P (x, y, z), then AP = (x-1, Y-2, Z-1), Pb = (- 1-x, 3-y, 4-z), from AP = 2PB, X-1 = 2 (- 1-x), x = - 1 / 3y-2 = 2 (3-y), y = 8 / 3z-1 = 2 (4-z), z = 3, so point P (- 1 / 3,8 / 3,3). PC = (4 / 3, - 5 / 3, - 2), | PC | = √ (77) / 3

Given that the equation of curve C is y = 4 - (X-2) ^ 2 (0 ≤ x ≤ 4), let the intersection point of curve C and X axis be a and B, and p be any point on curve C, find the vector PA * vector Pb
The range of value of
I'm looking for scope
The first question has been solved,
2. When 2

When x = 2, y max = 4; when x = 0, y = 0; when x = 4, y = 0
So: 0

Given the points P (0, m) and Q (0, - M), a straight line passing through point P intersects the curve CX ^ 2 = 4Y at two points a and B. if the vector AP = λ and the vector Pb (λ is a real number), the proof is given
Prove: vector QP ⊥ (vector QA - λ, vector QB)

Let me change my way of thinking, change the known condition vector AP = λ, vector Pb (λ is a real number) into vector QP - vector QA = λ (vector QB - vector QP), so we can get (1 + λ) vector QP = vector QA + λ vector QB. So the vector QP ⊥ (vector QA - λ vector QB) to prove vector QP *

Given that a certain point a (2,0) on the circle x + y = 4, B (1,1) is a point in the circle, p q is a moving point on the circle, the trajectory equation of the midpoint of the line AP is obtained
Second, if the angle PBQ = 90 ° to find the trajectory equation of the middle point of PQ, we need detailed steps. Thank you

(the following x ^ 2 represents the square of x) (1) find the trajectory equation of AP midpoint (x, y) XP = 2x-2, YP = 2yx ^ 2 + y ^ 2 = 4 (2x-2) ^ 2 + (2Y) ^ 2 = 4ap midpoint: (x-1) ^ 2 + y ^ 2 = 1 (2) if the angle PBQ = 90 °, find the trajectory equation of PQ midpoint (x, y) XP + XQ = 2x, (XP + XQ) ^ 2 = (2x) ^ 2

Given the fixed point a (4,0) and the circle M: x ^ 2 + y ^ 2 = 9 / 4, let B be the moving point on the circle m, and point P satisfy AP vector = 2PB vector,
(1) Find the trajectory equation of point P. (3) translate the point P obtained from (1) according to the vector a = (2 / 3,3) to get the trajectory C, make a tangent RT from a point R (x0, Y0) outside the trajectory C to the trajectory C, t is the tangent point, RT = RO, O is the origin, and find the minimum value of RT
Just ask the third question,

Let P coordinate (x, y) and B coordinate (3cosa / 2,3sina / 2), then: AP = (x, y) - (4,0) = (x-4, y) Pb = (3cosa / 2,3sina / 2) - (x, y) = (3cosa / 2-x, 3sina / 2-y), and: AP = 2PB, so: (x-4, y) = 2 (3cosa / 2-x, 3sina / 2-y), that is: X

If point P (0,2) to circle C (x + 1) ^ 2 + y ^ 2 = 1a is a moving point vector AB = 3 vector AP on circle C, then the trajectory equation of point B is?

Let B (x, y)
A（x0,y0）
Vector AB = 3 vector AP
（x-x0,y-y0）=3(-x0,2-y0)
x-x0=-3x0
y-y0=6-3y0
x0=-x/2
y0=(6-y)/2
The above x0, Y0 can be substituted into the circle equation

The hyperbola passing through the origin has a focal point F (4,0), 2A = 2
The answer is two circles, but to cut out two points, I want to ask which two points are cut out?

Let the center of the hyperbola be (x, y). Then the other focus is (2x-4,2y). Because the curve passes through the origin, the absolute value of the distance difference between the origin and the two focuses is 2A = 2. That is | 4 - √ [(2X-4) ^ 2 + (2Y) ^ 2] | = 2. = = = = > | 2 - √ [(X-2) ^ 2 + y ^ 2] | = 1. = = = > (X-2) ^ 2 + y ^ 2 = 1, or (X-2) ^ 2 + y ^ 2 = 9. That is, the trajectory is two circles

In equilateral △ ABC, Bo and co bisect ∠ ABC and ∠ ACB respectively, and the vertical bisectors of Bo and co intersect BC at e and f respectively? Why?

Be = CF, the reason is: connect OE, of, ∵ de vertical bisection ob ∵ be = OE (the distance from the point on the vertical bisection line of the line segment to the point at both ends of the line segment is equal), similarly, of = CF, ∵ EBO = ∵ BOE, ∵ FCO = ∵ FOC, ∵ equilateral triangle ABC, ∵ ABC = ∵ ACB = 60 ° (the angles of the equilateral triangle are equal and 60 °) ∵ Bo bisection ∵ ABC, CO bisection ∵ ACB ∵ EBO = 12 ∵ ABC = 30 ° and ∵ FCO = 12 ∠ AC B = 30 °, BOE = ∠ EBO = 30 °, FOC = ∠ FCO = 30 °, OEF = ∠ BOE + ∠ EBO = 60 °, ofe = ∠ FOC + ∠ FCO = 60 °, OEF is an equilateral triangle (a triangle with two inner angles of 60 ° is an equilateral triangle), OE = of = ef (all sides of an equilateral triangle are equal), be = EF = FC, i.e. be = CF

From a point Q on the hyperbola x2-y2 = 1, the perpendicular of the straight line x + y = 2 is drawn. If the perpendicular foot is n, then the trajectory equation of the midpoint P of the line QN is______ ．