Given that the moving point P of F1 (- radical 3,0) F2 (radical 3,0) satisfies | Pf1 | + | PF2 | = 4, find the maximum and minimum of vector Pf1 * vector PF2

Given that the moving point P of F1 (- radical 3,0) F2 (radical 3,0) satisfies | Pf1 | + | PF2 | = 4, find the maximum and minimum of vector Pf1 * vector PF2

The locus of point P is an ellipse with F! And F2 as the focus, C = √ 3, a = 2, and the elliptic equation is x ^ 2 / 4 + y ^ 2 = 1
Using the parameter equation of ellipse, assuming that the coordinates of point P are (x, y), the vector Pf1 = (- X - √ 3, - y), PF2 = (- x + √ 3, - y), Pf1 * PF2 = x ^ 2-3 + y ^ 2 = (4-4y ^ 2) - 3 + y ^ 2 = 1-3y ^ 2. The range of Y is [- 1,1], so the maximum value of Pf1 * PF2 is 1, and the minimum value is - 2

Given the points F1 (- 2,0) and F2 (2,0), the moving point P satisfies | PF2 | - | Pf1 | = 2. When the ordinate of point P is 12, the distance from point P to the origin of coordinate is ()
A. 62B. 32C. 3D. 2

It is known that a = 1, C = 2, B = 1, and the locus of point P is hyperbola x2-y2 = 1. Substituting y = 12, we can get x2 = 54, | op | = x2 + y2 = 54 + 14 = 62, so we choose a

Given the point F1 (- radical 2,0) F2 (radical 2,0), the ordinate of the point P satisfying the condition | PF2 | - | Pf1 | = 2 is 1 / 2, then the distance from the point P to the origin of the coordinate

If half focal length C = radical 2, | PF2 | - | Pf1 | = 2, then a = 1
The locus of P is the left branch of hyperbola
c^2=a^2+b^2
2=1+b^2
b^2=1
The equation is x ^ 2-y ^ 2 = 1

It is known that the locus of P satisfying the condition | PF2 | - | Pf1 | = 2 at two fixed points F1 (- √ 2,0) and F2 (√ 2,0) is e, and the line y = kx-1 intersects the curve e at two points a and B
(2) If | ab | = 6 √ 3, find the value of K

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)
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)
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
Because the value range of K is (- 2, - 1)
So k = √ 5 / 2