Given that the sum of the distances between the moving point P and the two foci F1 and F2 of hyperbola (x) 2 / 2 - (y) 2 / 3 = 1 is 6, the equation for finding the trajectory C of point P is obtained. If D (0,3), m and N are known to be on C and the vector DM = $vector DM, the value range of $is obtained Where did you copy it? It's totally wrong

Given that the sum of the distances between the moving point P and the two foci F1 and F2 of hyperbola (x) 2 / 2 - (y) 2 / 3 = 1 is 6, the equation for finding the trajectory C of point P is obtained. If D (0,3), m and N are known to be on C and the vector DM = $vector DM, the value range of $is obtained Where did you copy it? It's totally wrong

If the locus of hyperbola whose focal coordinates are (- radical 5,0) and (radical 5,0) P is an ellipse, then 2A = 6, a = 3, C = radical 5, then B ^ 2 = a ^ 2-C ^ 2 = 9-5 = 4, that is, the elliptic equation is x ^ 2 / 9 + y ^ 2 / 4 = 1 (1) when the line L is perpendicular to the X axis, PM = 1, PN = 5, then λ = PM / PN = 1 / 5 (2) when the line L is not perpendicular to the X axis

Given that the sum of the distances between the moving point P and the two foci F 1 and F 2 of hyperbola 2x-2y = 1 is 4, problem 1 is to find the equation of the locus C of the moving point P
If M is the moving point on the curve C, take M as the center and MF2 as the radius to make the circle M. if there are two intersections between the circle m and the Y axis, the abscissa value range of M can be obtained

1. Is the hyperbola x ^ 2-y ^ 2 = 1? If so, the solution is as follows: C = √ 2, focus coordinates F1 (- √ 2,0), F2 (√ 2,0). According to the conditions, its trajectory is an ellipse with the major axis of 4, which is confocal with the hyperbola, 2A = 4, a = 2, B ^ 2 = a ^ 2-C ^ 2 = 4-2 = 2, and the elliptic equation is: x ^ 2 / 4 + y ^ 2 / 2 = 1.2, let m coordinate (x0, Y0), if

If the hyperbola X29 − y2 = 1 has moving points P, F1 and F2, then the trajectory equation of the center of gravity m of △ pf1f2 is______ ．

A = 3, B = 1, C = 10, F1 (- 10, 0), F2 (10, 0) can be obtained from hyperbolic equation. Set point P (m, n), then & nbsp; m29 − N2 = 1. Let g (x, y) be the center of gravity of △ pf1f2, then x = m + 10 − 103, y = n + 0 + 03 can be obtained from the coordinate formula of the center of gravity of triangle

Given two fixed points F1 (- radical 2,0) F2 (radical 2,0), the length of the moving point P satisfies the condition that the length of PF2 - the length of Pf1 = 2, and the trajectory of point P is
Given two fixed points F1 (- radical 2,0) F2 (radical 2,0), the moving point P satisfies the condition that the length of PF2 - the length of Pf1 = 2, and the trajectory of point P is a curve e. the line L: y = kx-1 intersects with the curve e at two points a and B, if the length of AB = 6, radical 3, if there is a point C on the curve e, it is a vector OA + vector ob = m, vector OC, and the value of real number m is obtained
2. Given n (radical 5,0), P is the moving point on the circle M: (x + radical 5) ^ 2 + y ^ 2 = 36, the vertical bisector l of line PN intersects pm at point Q (1), and the equation for finding the locus C of point q is obtained
(2) If the line y = x + m intersects the curve C at two points a and B, the maximum area of the triangle AOB is obtained

1) The trajectory of P is the left branch of the hyperbola focusing on F1 and F2
And C = √ 2, a = 1
The equation of E is x ^ 2-y ^ 2 = 1 (x ≤ - 1)
2) Using the idea of combination of number and shape, a straight line passes through a fixed point (0, - 1) with a slope of K
According to the fact that there are two intersections between the straight line and the curve e, and K = - 2, the straight line and the curve are tangent,
The value range of K is (- 2, - 1)
3) If x ^ 2-y ^ 2 = 1 and y = kx-1 are combined, then (1-k ^ 2) x ^ 2 + 2kx-2 = 0
Let: a (x1, Y1), B (X2, Y2)
So X1 + x2 = 2K / (k ^ 2-1), x1x2 = 2 / (k ^ 2-1)
|AB|=[√(k^2+1)]|x1-x2|=6√3
The solution is k ^ 2 = 5 / 4 or 5 / 7
From (1), the value range of K is (- √ 2, - 1)
So k = √ 5 / 2
Point C is the intersection of the line passing through the origin O and the midpoint of the line segment AB and the curve E
The coordinates of the midpoint of line AB are m (- 2 √ 5,4)
So C (- √ 5,2), M = 2
The area of triangle ABC is s = 5 √ 3