The right focus f (C, 0) and point a (a ^ 2 / C, 0) of the ellipse. If there is a point P on the ellipse that satisfies the vertical bisector of line AP passing through point F, then the eccentricity of the ellipse is 0

The right focus f (C, 0) and point a (a ^ 2 / C, 0) of the ellipse. If there is a point P on the ellipse that satisfies the vertical bisector of line AP passing through point F, then the eccentricity of the ellipse is 0

According to the meaning of the question, if the distance between P and F on the ellipse is equal to | AF | then the maximum distance between P and F on the ellipse should be greater than | AF | and the maximum distance is a + C, | AF | = A & # 178 / / C-C ｜ a + C > A & # 178 / / C-C ｜ AC + C & # 178; > A & # 178; - C & # 178;; 2C & # 178; + AC-A & # 178; > 0, divide both sides by a & # 178; 2

Two triangles with equal base and height may not form a parallelogram______ (judge right or wrong)

Two triangles with equal base and height only have the same area, but their shapes are not necessarily the same. Two triangles that are not exactly the same cannot be combined into a parallelogram

The eccentricity of the ellipse x 2A2 + y 2B2 = 1 (a > b > 0) is 32. The ellipse and the straight line x + 2Y + 8 = 0 intersect at P, Q, and | PQ | = 10. The equation of the ellipse is obtained

From C2 = A2-B2, we get A2 = 4B2. From x24b2 + y2b2 + y2b2 + y2b2 = 1x + 2Y + 8 = 0, eliminate x, get 2y22 + 8y + 16-b2 = 0. From the root and coefficient relationship, we get Y1 + y2 = -4, y1y2 = 16 − B22. PQ PQ | 2 = A2 = 4B2 = 4b2.from C2 = A2 = 4B2. From the relationship of root and coefficient, we get Y1 + Y2 + Y2 + Y22 = - 4, Y1 + Y1 + 2 = 2 (x2-x1) 2 + (y2-y1-y1-y1) 2) 2 = 5 (y1-y1-y2-y2) 2) 2) 2) 2 = 5 (y1-2) 2) 2) 2 = 5 (y1-2) 2) 2) 2) 2 = 5 (y1-2) 2) 2) 2 = 5 [(1-2) 2 it is x236 + Y29 = 1

If the perimeter of a rectangle is 4a-b and one side is 2a-b, the other side is 2a-b-----
Don't forget that a rectangle is two long and two wide

"Mathematical Q & a group" answers for you, hoping to help you. [4a-b - (2a-b) × 2] △ 2 = B / 2, then the other side is B / 2. If you agree with my answer, please accept it in time, ~ if you agree with my answer, please click the [accept as satisfactory answer] button in time ~ ~ friends who ask questions on the mobile phone comment on the top right corner of the client

Given that the radius of ⊙ o is 1, PA is the tangent of ⊙ o, a is the tangent point, PA = 1, chord AB = root 2, find the length of Pb
There is no picture for this question

If you draw a diagram, you can see that there are two possibilities for the location of AB, as shown in the figure:
I think you have no problem with the calculation

Simplify the detailed process of finding the vector NQ + QP + mn-mp,

NQ+QP+MN-MP
=(NQ+QP)+MN-MP
=NP-MP+MN
=(NP+PM)+MN
=NM +MN
=0

In the polyhedral ABCDEF, the quadrilateral ABCD is a square with side length of 1, and the triangle ade and BCF are regular triangles, EF is parallel to AB, EF = 2, and the volume of the sphere

Draw a picture, EF parallel AB, polyhedron can be cut into three parts, take eg = 0.5, FH = 0.5, GH is 1, abcdgh is a column with bottom area of 0.5, height of 1, the other two parts are the same volume of vertebral body, calculate a volume multiplied by 2, vertebral bottom area of 0.5, height of 0.5

The known set a = {x MX + 1 = 0}, B = {x x}

mx + 1 = 0
When m = 0, X has no solution and a is an empty set
When m ≠ 0, x = - 1 / m
Because a ∩ B ≠ Φ
So - 1 / M < 0
So m > 0

Through a point P (m, 0) on the positive half axis of X axis, make a straight line L intersection ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1 and two points a and B, vector AP = 2Ab, find the range of M

The use of super staring out of the number of ability
Let: a (x1, Y1)
B（X2,Y2）
Then the linear equation of a, B and P is Y-Y1 = (y1-y2) / (x1-x2) × (x-x1)
When y = 0
Y1＝（Y2－Y1）/（X1－X2）×（X－X1）
X＝（X1－X2）×Y1/（Y2－Y1）＋X1
Because AP = 2Ab
So the absolute value of Y1 = 3 times the absolute value of Y2
Take it into the equation about X
When Y1 × Y2 > 0
X＝5/2X1-3/2X2
The maximum value of X is 12 (- 12)
When Y1 × Y2

13x12 = 13X () + () x2 = 12x () x () X3 = ()

13×12=13×(10)+(13)×2=12×(10)+(12)×3