In the plane rectangular coordinate system, O is the origin of the coordinate. Two points a (3, 1) and B (- 1, 3) are known. If point C satisfies the OC=α OA+β OB, where α, β∈ R and α + β = 1, find the locus of point C and its trajectory equation

In the plane rectangular coordinate system, O is the origin of the coordinate. Two points a (3, 1) and B (- 1, 3) are known. If point C satisfies the OC=α OA+β OB, where α, β∈ R and α + β = 1, find the locus of point C and its trajectory equation

C point satisfies
OC=α
OA+β
According to the collinear vector theorem, a, B and C are collinear
The locus of point C is a straight line ab
A (3, 1), B (- 1, 3),
The equation of line AB is y − 1
3−1＝x−3
From − 1 − 3, x + 2y-5 = 0
Therefore, the trajectory equation of point C is x + 2y-5 = 0

In the plane rectangular coordinate system xoy, if the curve X = If 4 − Y2 and the straight line x = m have and only one common point, then the real number M=______ .

The curve X =
4 − Y2 is the semicircle with origin o (0, 0) as the center and 2 as the radius (right side of Y axis)
There is and only one common point with the line L: x = m (L ∥ Y axis)
∴m=2
So the answer is 2

In the plane rectangular coordinate system xoy, the image of the first order function y = negative third root sign 3 + 3 intersects point a with X axis, B with y axis, and points D and E are points on line segments OB and ab respectively. When △ OAB is folded along De, point B falls on C on the edge of OA, and there is EC ⊥ Ao (1) Find the length of ab (2) Determine the shape of bdce and prove it (3) Find the analytic expression of De at this time

From y = - X / √ 3 + 3, we can see that ∠ OAB = 30 ° and ∠ B = 60 °
(1) Let x = 0 and y = 0 respectively, the coordinates of a and B are (√ 3,0), (0,1)
So AB = 2
(2)
Since [B=] [DCE=60], [OCD=30], [DC] [AB],
Because CE ‖ OB and be = EC,
So the quadrilateral bdce is rhombic
(3) It can be seen from the above that ∠ BDE = 60 ° so the angle between de and X axis is 30 °,
The analytic formula of De is: y = x / √ 3 + B
Let BD = a, then CE = be = a, AE = 2-a
From similar triangles:
CE/OB=AE/AB
That is: A / 1 = (2-A) / 2
3 / A is obtained
So the coordinates of point D are (0,1 / 3)
So the analytic formula of De is: y = x / √ 3 + 1 / 3

In the plane rectangular coordinate system xoy, it is known that the line L1 passes through point a (- 2,0) and point B (0,2 / 3 root sign 3), In the plane rectangular coordinate system xoy, we know that the line L1 passes through point a (- 2,0) and point B (0,2 / 3 root sign 3). The function analytic formula of line L2 is y = - radical 3 / 3x + 4 / 3 root sign * 3. L1 and L2 intersect with point P. circle C is a moving circle and its center C moves on line L1. Let the abscissa of center C be a vertical x-axis of CM, and the perpendicular foot is point M, (1) Fill in the blank: what is the function formula of line L1? What is the coordinate of point P? What is the degree of < EPB? (2) When circle C is tangent to line L2, please prove that the distance from point P to line cm is equal to the radius r of circle C, and write the value of a when r = 3 root sign 2-2 (3) When the circle C and the straight line L2 are not separated, the radius of circle C is known to be r = 3, and the root sign is 2-2. Note that the area of the quadrilateral nmop is s (where the point n is the intersection of the line cm and L2.) is there a maximum value of S? If there is, find the maximum value and the value of a at this time; if not, please explain the reason. PS diagram is drawn according to the meaning of the title. If it has not been drawn, Q I 576429388

It is easy to find the equation of the line y = genhao3 / 3x + 2genhao3 / 3, and the coordinate of point P is (1, genhao3). Because the position of point E is uncertain, the angle can not be calculated. If the circle C is tangent to the line L2, the distance from the center of the circle to L2 is r, and the coordinate of the center of the circle is (a, root sign 3 / 3A + 2 root sign 3 / 3)

In the plane rectangular coordinate system xoy, it is known that the segment length of the circle P cut on the X axis is twice the root sign 2, and the segment length on the Y axis is 2 times the root sign 3 In the plane rectangular coordinate system xoy, it is known that the segment length of circle P on the X axis is 2, and that on the Y axis is 2. 3; (2) If the distance from point P to the line y = x is (root 2) / 2, find the equation of circle P

(1) Let the center of the circle be p (a, b) and the radius R
R^2-b^2=2
R^2-a^2=3
So B ^ 2-A ^ 2 = 1
So the trajectory equation of the center P is y ^ 2-x ^ 2 = 1
(2)
R^2-b^2=2
R^2-a^2=3
|b-a|=1
The solution is: a = 0, B = 1, R ^ 2 = 3
②a=0,b=-1,R^2=3
So there are two such circles
X ^ 2 + (Y-1) ^ 2 = 3 and x ^ 2 + (y + 1) ^ 2 = 3

In the plane rectangular coordinate system xoy, it is known that ⊙ m passes through the points F1 (0, - C), F2 (0, c) and a (radical 3 * C, 0), where C > 0

By substituting F2 and a into the solution, M = (radical 3) C / 3, r = 2 (radical 3) C / 3, so the equation is [x - (radical 3) C / 3] + y = 4C / 3

In the plane rectangular coordinate system xoy, we know that the left and right focal points of the ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 are F1 and F2 respectively The focal length is 2, the equation of a directrix is x = 2, P is a point on the ellipse, the straight line Pf1 intersects ellipse C and another point Q (1) Find the equation of elliptic C; (2) If the coordinates of point P are (0, b), the equation of circle P, Q and F2 is solved; (3) If the vector f 1p = μ * vector QF1 and μ∈ [1 / 2,2], find the maximum value of vector OP * vector OQ This is mainly the third question,

Change the input in the graph to u
The answer is as shown in the figure, friendly prompt: click on the picture to view the large picture

In the plane rectangular coordinate system xoy, the left and right focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) are F1 (- C, 0), F2 (C, 0), known (1, e) and (e, √ 3 / 2) are all on the ellipse, where e is the eccentricity of the ellipse, then the equation of the ellipse is ()

X ^ 2 / 2 + y ^ 2 / 1 = 1, the reason (1, e) and (E, √ 3 / 2) are on the ellipse, e = C / A, i.e. 1 ^ 2 / A ^ 2 + e ^ 2 / b ^ 2 = 1 ^ 2 / A ^ 2 + C ^ 2 / A ^ 2B ^ 2 = 1, and then B ^ + C ^ = a? B? E ^ 2 / A ^ 2 + (√ 3 / 2) ^ 2 / b ^ 2 = C ^ 2 / A ^ 2A ^ 2 + 3 / 4B ^ 2 = 1

As shown in the figure, in the plane rectangular coordinate system xoy, the left and right focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) are F1 (- C, 0), F2 (C, 0). It is known that (1, e) and (E, √ 3 / 2) are on the ellipse, where e is the eccentricity of the ellipse (1) I have worked out the equation of ellipse by myself, mainly the second question (2) Let a and B be two points on the ellipse above the x-axis, and the line AF1 is parallel to the line BF2, and af2 and BF1 intersect at the point P ① If af1-bf2 = √ 6 / 2, calculate the slope of line AF1 ② Verification: Pf1 + PF2 is a constant value

1. This question can be called the most difficult elliptic question in Jiangsu college entrance examination in the past five years. 2. The two small questions in the second question of this question are closely related. 3. Experts provide many solutions, but the key is similarity. Similarity is also one of the difficulties
The second and third points are the key. I hope it can help you

1.1 as shown in Fig. 1, in the plane rectangular coordinate system, the straight line AB intersects the x-axis at point a, the intersection Y-axis with the point B, and the point C is the moving point on the straight line ab. (1) if ∠ OAB is 20 ° larger than ∠ oba, OC ⊥ AB, calculate the degree of ⊥ AOC. (2) as shown in Fig. 2, am bisection ∠ Bao, BM bisection ∠ OBN, when point a moves on the negative half axis of x-axis, Whether the value of ∠ AMB has changed? If not, find out the degree of ∠ AMB; if so, please explain the reason; (3) place two mirrors along AB and ob, and the reflected light DF and incident light OP intersect at point E. if ∠ OAB = 45 °, the following two conclusions are obtained: (1) DF / / AB; (2) DF ⊥ Op One and only one conclusion is correct. Please point out the correct conclusion and explain the reason

(1) If the reading of ∠ oba is x, then ∠ OAB = x + 20 °
∵∠OBA+∠OBA=90°
That is, x + 20 ° + x = 90 °
That is, x = 35 degrees
∴∠OAB=35°+20°=55°
OC ⊥ ab
∴∠OCA=90°
∴∠ AOC=90°-∠ OAB=90°-55°=35°