On conic curve The ellipse 1 / 2 x ∧ 2 + y ∧ 2 = 1 and a point m (0,2) outside the ellipse are known. Passing through this point leads to the trajectory equation of a straight line intersecting with the ellipse at the midpoint P of a and B

On conic curve The ellipse 1 / 2 x ∧ 2 + y ∧ 2 = 1 and a point m (0,2) outside the ellipse are known. Passing through this point leads to the trajectory equation of a straight line intersecting with the ellipse at the midpoint P of a and B

Let a (x1, Y1) B (X2, Y2) P (x0, Y0) so x0 = (x1 + x2) / 2y0 = (Y1 + Y2) / 2a, B on ellipse so X1 ^ 2 / 2 + y ^ 2 = 1x2 ^ 2 / 2 + Y2 ^ 2 = 1 subtract so (x1 + x2) (x1-x2) / 2 + (Y1 + Y2) (y1-y2) = 02x0 (x1-x2) + Y0 (y1-y2) = 0, so AB slope k = (y1-y2) / (x1-x2) = 2x0 / y

A high school mathematics problem (conic)
Let F 1 and F 2 be the left focus of ellipse e: X & sup2 / / A & sup2; + Y & sup2 / / B & sup2; = 1 (a is greater than B is greater than 0), the line L passing through F 1 with slope 1 intersects E and B, and the lengths of line segments af2, AB and BF2 form an arithmetic sequence
(1) Find the eccentricity of e
(2) Let the point P (0, - 1) satisfy the phase of the line PA and Pb, then the equation of the ellipse e is obtained

Yes, I do, but it's a lot of trouble to input a bunch of mathematical symbols here
Ask the experts. They are patient
Let me give you an idea
one
Let a (x1, Y1) B (X2, Y2)
Let the point oblique form of the line l be combined with the equation to write to WIDA's theorem, and then the first definition is used to solve it in the relation of arithmetic sequence
two
According to the meaning of the question, we can get P on the vertical bisector of ab. then we can get the satisfying relation of X1 and X2 according to the first question

A high school mathematics problem about conic curve
It is known that the center of the ellipse is the coordinate origin o, the intersection point is on the x-axis, the slope is 1, and the line L passing through the right focus F of the ellipse intersects the ellipse at two points a and B, and the vector OA + vector ob is perpendicular to the vector n = (1,3)
1. Find the eccentricity e of ellipse
2. Let m be any point on the ellipse, and the vector om = (M + n) vector OA + (m-n) vector ob (m, n belong to R), find the trajectory of the moving point n (m, n)

Find the limit of LIM (n drive near infinity) [ln (1 + 1 / N)] / [sin2 / N]

How does 1 / N (sin0 / 1 sin1 / 2 sin2 / 3... Sinn-1 / N) become ∫ (01) Si
xdx

This is the definition of definite integral (1 / N) {sin (0 / 1) + sin (1 / N) +... + sin [(n-1) / N]} = ∫ (0 - > 1) SiNx DX

lim(1/n^2+4/n^2+7/n^2+… +3n-1/n^2)

Sequence 1 + 4 + +The sum of 3n-1
Sn=n+3n(n-1)/2
=n+3n/2-3n/2
=3n/2-n/2
lim(1/n^2+4/n^2+7/n^2+… +3n-1/n^2)
=lim(3n^2-n/2n^2)
=lim(3-1/n)/2
=3/2

Given that the sequence {an} satisfies an = [LG (n + 2)] / [LG (n + 1)] n ∈ n *, the product A1 * A2 * ~ ~ ~ ~ * AK is defined as an integer, and K is called an expected number, then [12010] / [LG (n + 1)] n ∈ n *
Detailed process, thank you

An = [LG (n + 2)] / [LG (n + 1)] A1A2... AK = (Lg3 / LG2) (LG4 / Lg3)... [LG (K + 2) / LG (K + 1)] = [LG (K + 2) / LG2], the result of product is integer, only K + 2 is the power of 2. Let K + 2 = 2 ^ m, 1 ≤ K ≤ 20103 ≤ K + 2 ≤ 20123 ≤ 2 ^ m ≤ 20122 ≤ m ≤ 10, there are 9. Then k = 2,6,14,30,62

The known set a = {A1, A2 , AK (K ≥ 2)}, where AI ∈ Z (I = 1, 2 Two corresponding sets are formed by the elements in a: S = {(a, b) | a ∈ a, B ∈ a, a + B ∈ a}, t = {(a, b) | a ∈ a, B ∈ a, A-B ∈ a}. Where (a, b) is a pair of ordinal numbers, and the number of elements in sets s and T are m and N respectively. If there is always - a ∉ a for any a ∈ a, then set a is said to have property P. (I) test whether the sets {0, 1, 2, 3} and {- 1, 2, 3} are identical For the set with property P, write the corresponding set s and T; (II) for any set a with property P, prove that n ≤ K (K − 1) 2; (III) judge the size relationship between M and N, and prove your conclusion

(1) The set {0, 1, 2, 3} does not have the property P. the set {- 1, 2, 3} has the Property P, and its corresponding sets s and T are s = (- 1, 3), (3, - 1), t = (2, - 1), (2, 3). (II) prove: firstly, there are K2 ordinal number pairs (AI, AJ) composed of elements in a

With the Property P: for any I, J (1 ≤ I ≤ J ≤ n), at least one of AJ + AI and AJ AI is a term in the sequence
Known sequence a 2,..., a 2 ,an（0≤a1＜a2＜… For any I, J (1 ≤ I ≤ J ≤ n), at least one of AJ + AI and AJ AI is a term in the sequence
① The sequence 0,1,3 has the property P;
② The sequence 0,2,4,6 has the property P;
③ If sequence a has property P, then A1 = 0;
④ If the sequence A1, A2, A3 (0 ≤ A1 < A2 < A3) has the Property P, then a1 + a3 = 2A2
Among them, the true proposition has ②, ③ and ④
How should I write it down,

Interpretation: take any two items (can be the same) in the increasing sequence, and at least one of their sum or difference is one of the items in the sequence
1 is wrong, because the difference between 1 and 3, 2 and 4, are not in the sequence
For 2 pairs, the sum or difference of 0 and any number is the term of the sequence, the sum of 2 and 4, the difference of 6, and the difference of 4 and 6 are the term of the sequence
If we take the largest term twice, the sum of them cannot be in the sequence, so the difference 0 must be the term of the sequence
4 pairs, the same as above, A1 = 0, because A2 > 0, so A2 A3 > A3 does not belong to the sequence, so the difference belongs to the sequence, may be A2 or A3 (A2 = 0 rounding), so it can only be a3-a2 = A2

Which theorem of sequence can be used to deduce this formula? A1 + an = A2 + an-1 = A3 + An-2 = =ak+an-k+1,k∈{1,2,… ,n}

Is to use the general formula of arithmetic sequence
a1+an=a1+a1+(n-1)d=2a+(n-1)d
a2+a(n-1)=a1+d+a1+(n-2)d=2a1+(n-1)d
a3+a(n-2)=a1+2d+a1+(n-3)d=2a1+(n-1)d
.
ak+a(n-k+1)=a1+(k-1)d+a1+(n-k)d=2a1+(n-1)d
So, a1 + an = A2 + an-1 = A3 + An-2 = =ak+an-k+1,k∈{1,2,… ,n}