It is known that the focus of ellipse is F1 (0, - 1), F2 (0,1), and eccentricity e = 1 / 2 Let p be on the ellipse, and | Pf1 | - | PF2 | = 1, and find the value of cos ∠ f1pf2

It is known that the focus of ellipse is F1 (0, - 1), F2 (0,1), and eccentricity e = 1 / 2 Let p be on the ellipse, and | Pf1 | - | PF2 | = 1, and find the value of cos ∠ f1pf2

e=1/2
c=1 a=2
|PF1|+|PF2|=4 |PF1|=2.5 |PF2|=1.5
COS∠F1PF2=(2.5^2+1.5^2-2^2)/2*2.5*1.5=0.6

It is known that the two focal points of ellipse e are F1 (- 1,0) and F2 (1,0), and its eccentricity e = 1 / 2. The equation for solving ellipse e

In an ellipse, if C = 1, e = A / C = 1 / 2, a = 2 and B = root 3 are known, then the elliptic equation is, (x ^ 2) / 4 + (y ^ 2) / 3 = 1

Given three points P (5,2), F1 (- 6,0), F2 (6,0), 1. Find the standard equation of ellipse with F1, F2 as focus and passing through point P. 2. Set point P, f1.f2
On the standard equation of hyperbola with symmetric point share of P ', F 1', F 2 'as focal point and passing through p' of line y = X

It is known that the ellipse e passes through points (2,3), the axis of symmetry is the coordinate axis, the focal points F1 and F2 are on the x-axis, and the eccentricity e = 1 / 2 (1) to solve the equation of the ellipse E; (2) to solve the angle F

I found the original question and answer on the Internet. I don't know if it's what you want
It is known that the ellipse e passes through point a (2,3), the axis of symmetry is the coordinate axis, the focus F1 and F2 are on the X axis, and the eccentricity e = 1 / 2
(1) Find the equation of ellipse E;
(2) Find the equation of the line L where the bisector of the angle f1af2 is located;
(3) Are there two different points about the symmetry of line L on the ellipse e? If so, please find out; if not, explain the reason
(1) The equation of ellipse with focus on x-axis and eccentricity of 1 / 2 can be set as x2 / A2 + 4y2 / 3a2 = 1
If we take x = 2 and y = 3, we get a = 4, so the elliptic equation is x2 / 16 + Y2 / 12 = 1
(2) The focus is F1 (- 2,0), F2 (2,0). Obviously, that is to say, f1af2 is a right triangle with 345 sides
If there is a point on the line which is located in the interior of the triangle f1f2a and the distance to the x-axis is 1, then the point must be the center of the triangle. The point is easy to find m (1,1), so the line L where the angular bisector of f1af2 is is the line am, and the equation is easy to find L: y = 2x-1
(3) Suppose that such two points exist, then the slope of the straight line passing through two points is - 1 / 2, let the linear equation be y = - 1 / 2x + B, and it is combined with the elliptic equation to get x2-bx + b2-12 = 0. Because there are two different intersections, the discriminant of the equation is greater than zero, that is, b2-4 (b2-12) > 0, then - 4 is obtained

It is known that F1 and F2 are the left and right focus of ellipse x square / a square + y square / b square,
There is a point P on the ellipse, so that ∠ f1pf2 = 90 degrees, then the value range of the eccentricity of the ellipse

Let Pf1 PF2 = 0, then the trajectory equation of the point P satisfying the condition is x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\; -

It is known that the square of x square / A + the square of Y square / b of the ellipse is 1, f1.f2 is its focus, P is any point of the ellipse, and the value range of Pf1 · PF2 is [2.3]. Let the left and right vertices of the ellipse be A.B respectively, and l be the right guide line and the straight line of the ellipse PA.PB Intersect l at two points m.n, find vector MF1 and vector MF2

Set the coordinates of point P, then use the elliptic equation to get the equation of L, and use the deviation rate and the definition of the collimator to get it

It is known that the ellipse C: the square of X / the square of a + the square of Y / the square of B = 1 (a > O, b > o). F1 and F2 are the two focal points of ellipse C,
If point P is a point on the ellipse and satisfies / PF2 / = / F1F2 /, and the distance from F2 to the straight line Pf1 is equal to the length of the minor axis of the ellipse, the eccentricity of the ellipse is calculated?

Point P is on the elliptic C, and 124; Pf1 | 124124;pf1 | |pf2 124pf2 124\124| F1F2, you can get D as D is the midpoint of Pf1, D is the midpoint of Pf1, D is the midpoint of Pf1, so DF1, so DF1 is| df2 | 2 = |

It is known that the eccentricity of ellipse C: square of X / square of a + square of Y / square of B = 1 (a > b > 0) is 1 / 2, and the left and right focal points are F1, F2 respectively,
If point G is on the ellipse and vector GF1 × GF2 = 0, the area of △ gf1f2 is 3
1. Solving the equation of ellipse C
2: Let the left and right vertices of the ellipse be a and B. the line L passing through point F2 intersects the ellipse with two different points m and n (different from points a and b). Explore whether the intersection of lines am and BN can be on a fixed line perpendicular to the X axis. If you can request this fixed line, please explain the reason

I will write the first question simply, mainly the second one
1.GF1*GF2 = 2*3=6 ; GF1^2+GF2^2 = F1F2^2 ; 4a^2-2*6 = 4c^2
So a = 2, C = 1, B = √ 3 equation x ^ 2 / 4 + y ^ 2 / 3 = 1
2. The key to this question is to set but not to ask, and to deal with it calmly. Don't be confused. Look at me
Let Mn: X-1 = my, am: x + 2 = m1y, BN: X-2 = M2y
Points m (x1, Y1); n (X2, Y2)
&Mining the known conditions as far as possible:
Finding the intersection point K of Ma, Nb
The results show that: simultaneous x + 2 = m1y; X-2 = M2y; easy to know y = 4 / (m1-m2); X = 2 (M1 + m2) / (m1-m2) = 2 + 4m2 / (m1-m2). 1#
Simplify the question "can the intersection of lines am and BN be on a fixed line perpendicular to the x-axis? If you can request this fixed line, please explain the reason." this is equivalent to proving whether x = 2 + 4m2 / (m1-m2) is a fixed value
M is the intersection of Mn and am, and N is the intersection of Nb and Mn
X1 + 2 = m1y1; X1 - 1 = my1; so M1 = m + 3 / Y1
X2-2 = m2y2; x2-1 = my2; so M2 = M-1 / Y2 into 1#
x = 2+4 * [m-1/y2]/[3/y1+1/y2] = 2 + 4* [ my1y2-y1]/[3(y1+y2)-2y1].2#
Line Mn: my = X-1 into ellipse 3x ^ 2 + 4Y ^ 2-12 = 0
It is known that (3m ^ 2 + 4) y ^ 2 + 6my-9 = 0
So my1y2-y1 = - 9m / (3m ^ 2 + 4) - Y1
3(y1+y2)-y1 = -18m/(3m^2+4)-2y2
Obviously (my1y2-y1) / (3 (Y1 + Y2) - Y1) = 1 / 2
SO 2 # = 2 + 4 * (1 / 2) = 4
That is to say, x = 4 is a fixed value, so there is a straight line, that is, x = 4

It is known that the eccentricity of the ellipse C x squared / a squared + y squared / b squared = 1 (a > b > 0) is half, and the left and right focal points are F1 and F2 respectively,
Point G is on the ellipse and the area of vector GF1 × GF2 = 0 △ gf1f2 is 3
1. Solving the equation of ellipse C
2. If the line L and the ellipse intersect at two points P O, and the vector op · OQ = 0, O is the origin, explore whether the distance between the point O and the line L is a fixed value, if the fixed value is not a reason

(1) Let | f1f1f2 = 2C (c > 0) then F1 (- C, 0), F2 (C, 0) \124; f (F, f (F, F, f (C, 0) \124; f | f | (f | f | (f | f | (f | (f | (f | f | (f | (f | = 2C = 2C = 2C = 2) (C | (C = f | = 2C = 2C = 2C (2) | (C = f | = 2C = 2C = 2) | (C = 2C = 2) | (C = 2C = 2C = 2C = 2C = 2C = 2C = 2C = 2C = 2C (C = 2) | (C = 2C (C 178; - 2D1 * D

As shown in the figure, F1 and F2 are the left and right focus of ellipse x square / a square = 1 (a > b > 0). When the eccentricity value is in what range, there is always a point P on the ellipse, making Pf1 ⊥ PF2
The graph is an ellipse centered on the center of a circle, with the focus on the x-axis. A right triangle is right on the ellipse, and the two focuses are the bottom of the right triangle

If there is always a point P on the ellipse X & sup2; / A & sup2; + Y & sup2; / B & sup2; = 1 (a > b > 0) such that Pf1 ⊥ PF2 = 0, then the trajectory equation of the point P satisfying the condition is X & sup2; + y & sup2; = A & sup2; - B & sup2; ① and the elliptic equation x & sup2; / A & sup2; + Y & sup2; / B & sup2; = 1 ②