Let p be the point on the ellipse X & # 178 / A & # 178; + Y & # 178 / B & # 178; = 1 (a > b > 1), and the two focal points are F1 and F2 respectively If ∠ pf1f2 = 75 ° and ∠ pf2f1 = 15 °, what is the eccentricity of the ellipse?

Let p be the point on the ellipse X & # 178 / A & # 178; + Y & # 178 / B & # 178; = 1 (a > b > 1), and the two focal points are F1 and F2 respectively If ∠ pf1f2 = 75 ° and ∠ pf2f1 = 15 °, what is the eccentricity of the ellipse?

In the right triangle mf1f2,
MF1+MF2=F1F2cos15 º+F1F2sin15 º
=√2F1F2sin60º
According to the definition of ellipse, MF1 + MF2 = 2A, F1F2 = 2C,
2 a = radical 2 * 2C × √ 3 / 2,
That is, C / a = (√ 6) / 3,
The eccentricity of ellipse is (√ 6) / 3
cos(-17/6π)+tan(-17/6π)
cos(-17/6π)+tan(-17/6π)
=-cos(π/6)+tan(π/6)
=-√3/2+√3/3
=-√3/6
cos(-17/6π)+tan(-17/6π)
=cos(7/6π-4π)+tan(1/6π-3π)
=cos(7/6π)+tan(1/6π)
=-√3/2+√3/3
=-√3/6
It is known that F1 and F2 are the two focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 (a > b > 0). If the circumference of △ af1b is 16, the ellipse will be formed
It is known that F1 and F2 are the two focal points of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 (a > b > 0). If the chord ab of the ellipse is made through F2, if the perimeter of △ af1b
It's 16, eccentricity of ellipse e = √ 3 / 2
(1) Find the standard equation of ellipse;
(2) If the angle f1af2 = 90 °, calculate the area s of △ f1af2
(3) It is known that P (2,1) is a point in the ellipse. Find a point Q on the ellipse so that √ 3pq + 2qf & ᦇ 61483; is minimum, and find the minimum value
Help to solve (2) (3) two questions
If the equation of line L is uncertain, it may lead to the uncertainty of elliptic equation
It is easy to know that the left quasilinear is x = - A ^ 2 / C, O (0,0), F2 (- C, 0)
Let the equation of line l be y = KX + m (K ≠ 0), P1 (x0, Y0)
Let the symmetric point of O with respect to the line l be Q
From the definition of ellipse, we know that p1f1 + p1f2 = 2A
It is known that p1f2-p1f1 = 10A / 9
If the above two equations are added together, we can get the formula p114a
The distance formula between two points is p1f1 ^ 2 = (x0 + C) ^ 2 + Y0 ^ 2
So (x0 + C) ^ 2 + Y0 ^ 2 = (14a / 9) ^ 2 (1)
Because point P1 is on the ellipse, and notice that B ^ 2 = a ^ 2-C ^ 2
Then x0 ^ 2 / A ^ 2 + Y0 ^ 2 / (a ^ 2-C ^ 2) = 1 (2)
And point P is on line L
Then Y0 = kX0 + m (3)
Because o and Q are symmetric with respect to the line L, q is on the line passing through O and perpendicular to the line L
Notice that the slope of line L is K
Then let the equation of the line passing through O and perpendicular to the line l be y = - X / K
And Q is on the directrix = - A ^ 2 / C
Q (- A ^ 2 / C, a ^ 2 / KC) is obtained by solving the above two linear equations
Obviously, the line L is the vertical bisector of the segment OQ
Then the distance from P1 to O and Q is equal, that is, p1o = p1q
The formula of distance between two points is (x0 + A ^ 2 / C) ^ 2 + (y0-a ^ 2 / KC) ^ 2 = x0 ^ 2 + Y0 ^ 2
(2 / C) x0 - (2 / KC) Y0 + (a ^ 2 / C ^ 2) (1 + 1 / K ^ 2) (4)
If the line L is determined, that is, K and m are determined, a and C can be determined by using the above four equations, and then B can be determined, and finally the elliptic equation can be determined
A = 4. B = 2, C = 2 √ 3. Suppose AF1 = X. af2 = y. using the first cosine theorem, we can get X & # 178; + Y & # 178; = 48. According to the definition of ellipse, which F is x + y = 2A = 8  xy = 8, s = &# 189; xy = 4? Set the coordinates of point Q, establish the objective function and get the maximum value.
Calculate the following values (1) sin5 π / 2 + cos & # 178; 17 π / 3-tan & # 178; 23 π / 6
sin5π/2+cos²17π/3-tan²23π/6
sin5π/2+cos²17π/3-tan²23π/6
=sinπ/2+cos²π/3-tan²(-π/6)
=1+1/4-1/3
=11/12
F1, F2 ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) two focus, through F2 as elliptic string AB, if △ af1b circumference is 16, eccentricity is root sign 3 / 2, find the equation of ellipse
(AF1 + af2) + (BF1 + BF2) = 16, that is, 2A + 2A = 16
e＝c／a
c²＝a²＋b²
We can find a and B from above
The eccentricity of ellipse is less than 1. You are more than one. What's the matter? Let me tell you the method, the circumference of af1b = 4A, a = 4, eccentricity e = A / C, we can find C, then a ^ 2 = B ^ 2 + C ^ 2, and then substitute it.
Given cos (π + α) = 1 / 2, calculate sin (2 π - α) sin [(2n + 1) π + α] + sin [α - (2n + 1) π] / sin (2n π + α) cos (α - 2n π)
cos(π+α)=1/2
∴cosa=-1/2
∴sina=±√3/2
The original form
=[(-sina)(-sina)+(-sina)]/(sinacosa)
=(sina-1)/cosa
=2±√3
The original formula is sin (- α) sin (2n π + π + α) + sin (α - 2n π - π) / sin α cos α
=（-sinα）（-sinα）+（-sinα）/sinα cosα
=（sinα-1）/cosα
∵cos(π+α)=-cosα=1/2
∴cosα=-1/2
∴sinα=±√3/2
The original formula = 2 - √ 3... Expansion
The original formula is sin (- α) sin (2n π + π + α) + sin (α - 2n π - π) / sin α cos α
=（-sinα）（-sinα）+（-sinα）/sinα cosα
=（sinα-1）/cosα
∵cos(π+α)=-cosα=1/2
∴cosα=-1/2
∴sinα=±√3/2
The original formula is 2 - √ 3 or 2 + √ 3
[this question is about induction formula]
[the basic principle and process have been written in detail. If you don't understand, please ask. 】What about sin (2 π - α)
It is known that the left and right focal points of the ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1 are f 1F 2 respectively. A straight line L crosses the ellipse through F 1 at two points a and B
If the slope of L is 1, find the perimeter of triangle abf2
It's not perimeter, it's area dial the wrong number
The focus F1, F2 coordinates are easy to get (1,0) (- 1,0)
No matter which focus you pass, the area is the same
Let F 1 (1,0), then the equation of L is y = X-1
Let the coordinates of intersection be (x1, Y1) (X2, Y2)
In the elliptic equation
（y+1）²/4+y²/3=1
Its area = | F1F2 | (| Y1 | + | Y2 |) / 2, | F1F2 | = 2
It's obvious that | Y1 | + | Y2 | = | y1-y2|
Y1, Y2 are the univariate quadratic equations (y + 1) & sup2 / 4 + Y & sup2 / 3 = 1
According to Weida's theorem
y1+y2=-6/7
y1*y2=-9/7
So it's easy to get
|y1-y2|=√（（y1+y2）²-4y1y2）=12√2/7
So area = 12 √ 2 / 7
How to simplify the high order of Mathematics
cos(27+a)cos(33-a)-sin(27+a)sin(33-a)=cos[(27a+a)+(33-a)]=cos60=1/2
Do not know, welcome to ask
Integral sum difference formula
cosαcosβ=[cos（α+β）+cos（α-β）]/2
sinαsinβ=[-cos（α+β）+cos（α-β）]/2
The original formula is cos (α + β) = cos60 = 0.5
It is known that the left and right focal points of the ellipse CX ^ 2 / 9 + y ^ 2 / 8 = 1 are F1F2 respectively. Make a straight line intersection ellipse C through F1 at two points ab
1 find the maximum area of triangle abf2
2 calculate the value of tanf1af2 when the area of triangle abf2 reaches the maximum
(detailed process)
1. The maximum area is 16 / 3
A = √ 9 = 3, B = √ 8 = 2 √ 2, C = √ (A & # 178; - B & # 178;) = 1, so | F1F2 | 2C = 2
The linear equation passing through F1 is: x + 1 = ay (this is to consider the case that a = 0 is perpendicular to the X axis). Let the intersection of the equation and the ellipse a (x1, Y1), & nbsp; B (X2, Y2). It is obvious that Y1 and Y2 are different signs
S△ABF2=S△AF1F2+S△BF1F2
=|F1F2|*|y1|/2 + |F1F2|*|y2|/2
=|y1|+|y2|
=|y1-y2|
(AY-1) &# 178; / 9 + Y & # 178; / 8 = 1, and (8A & # 178; + 9) y & # 178; - 16ay-64 = 0
So Y1 + y2 = 16A / (8A & # 178; + 9), y1y2 = - 64 / (8A & # 178; + 9)
The function of 64x + 1 / X is the smallest when x = 1 / 8, and then increases with X. because a & # 178; + 1 & gt; 1, so after | a | increases, the denominator increases, so the area decreases. So when a = 0 (the line is perpendicular to the X axis), the area is the largest, equal to 16 / 3. The two points are a (- 1,8 / 3) and B (- 1, - 8 / 3)
2. tan F1AF2 = 3/4.
At this time, af1f2 is a right triangle, Tan & nbsp; f1af2 = | F1F2 | / | AF1 | = 2 / (8 / 3) = 3 / 4
Simplify (sin α - cos α) ^ 2
（1）(sinα-cosα)^2
（2）sin(θ/2)cos(θ/2)
I'm depressed. I've written more than N, which is different from the answer,
(1) Original formula = sin α ^ 2 + cos α ^ 2-2sin α cos α
=1-sin2α
(2) Original formula = 1 / 2Sin θ
Double angle formula sin (2a) = 2sinacosa
1，1-sin（2a）
2，（sina）/2
1-sin(2a)