Let the left and right focus of ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) be F1 and F2 respectively. Point P (a, b) satisfies PF2 = F1F2 (1) Find the eccentricity e of ellipse; (2) Let the line PF2 and the ellipse intersect at two points a and B. if the line PF2 and the circle (x + 1) ^ 2 + (y-radical 3) ^ 2 = 16 intersect at two points m and N, and Mn = (5 / 8) AB, the equation of the ellipse is obtained

Given sin (π 4 − x) = 35, the value of sin2x is ()
A. 1925B. 1625C. 1425D. 725
Method 1: from the known 22 (cosx − SiNx) = 35, the square of both sides is 12 (1 − sin2x) = 925, and then get sin2x = 725; method 2: let π 4 − x = α, then sin α = 35, so sin2x = sin (π 2 − 2 α) = Cos2 α = 1 − 2sin2 α = 725
The two focal points of ellipse C are F1 and F2, the point P is on ellipse C, and Pf1 ⊥ F1F2, | PF2 | = 4 / 3, | PF2 | = 14 / 3
Error correction, | Pf1 | = 4 / 3
For the convenience of writing, let m = | Pf1 | = 4 / 3, n = | PF2 | = 14 / 3, long axis 2a, focal length 2C
∵ is an ellipse, ∵ m + n = 2A, that is, a = (M + n) / 2 = 3, a & sup2; = 9
∵m⊥F1F2
In RT △ pf1f2, M & sup2; + (2C) & sup2; = n & sup2;, C & sup2; = 5
∴b²=a²-c²=4
The elliptic equation is X & sup2 / 9 + Y & sup2 / 4 = 1 or X & sup2 / 4 + Y & sup2 / 9 = 1
sin(45-x)=3/5,sin2x=?
sin(45°-x)=3/5
cos(45°+x)=3/5
The multiplication of two formulas sin (45 ° - x) cos (45 ° + x) = 9 / 25
Then (1 / 2) (sin90 ° - sin2x) = 9 / 25
sin2x=1-18/25=7/25
Sinxcosx can be obtained from the square of 7 / 25 and sin2x = 2sinxcosx
X45
sin(45-x)=3/5, cos(45-x)=-4/5 cos(90-2x)=sin(45-x)*sin(45-x)-cos(45-x)*cos(45-x)=7/25
sin2x=7/25
So sin2x = 7 / 25
Let t = 45-x
So sin (T) = 3 / 5
=>cos(2t)=1-2sin(t)^2=1-2*9/25=7/25
And COS (2t) = cos (90-2x) = sin (2x)
So sin (2x) = 7 / 25
Using angle substitution can avoid complex operation
sin(45-x)=3/5,cos(90-45+x)=sin(45-x)=3/5
That is cos (45 + x) = 3 / 5, Cos2 * (45 + x) = cos (90 + 2x) = - sin2x = 2cos ^ 2 (45 + x) -- 1 = - 7 / 25
That is sin2x = + 7 / 25
Let the left and right focal points of the ellipse x ^ 2 / 16 + y ^ 2 / b ^ 2 = 1 (4 > b > 0) be F1 and F2 respectively, and the point P (4, b) satisfies the equation of | PF2 | = | F1F2 | (1) to find the ellipse
(2) If the line PF2 intersects the circle (x + 1) ^ 2 + (y ^ radical 3) ^ 2 = 16 at two points m and N, find | Mn|
（1）|F1F2|=2c,|PF2|=√[(4-c)²+b²]=√[(4-c)²+a²-c²]；
According to the meaning 2C = √ [(4-C) &# 178; + A & # 178; - C & # 178;], 4C & # 178; = 4 & # 178; - 8C + A & # 178;;
Substituting a = 4 into the solution, C = 2; B & # 178; = 16-2 & # 178; = 12; so the elliptic standard equation is X & # 178 / 16 + Y & # 178 / 12 = 1;
(2) Coordinates F2 (2,0), P (4,2 √ 3); linear equation of PF2: y = √ 3 * (X-2);
The radius of circle (x + 1) &# 178; + (Y - √ 3) &# 178; = 16 is r = 4, and the center coordinates (- 1, √ 3);
The distance between circle center and straight line PF2 (chord center distance) d = | Y - √ 3 (X-2) | / 2 = | √ 3 - √ 3 (- 1-2) | / 2 = 2 √ 3;
∴ |MN|=2√(R²-d²)=2√[4²-(2√3²)]=4；
Given sin (π 4 − x) = 35, the value of sin2x is ()
A. 1925B. 1625C. 1425D. 725
Method 1: from the known 22 (cosx − SiNx) = 35, the square of both sides is 12 (1 − sin2x) = 925, and then get sin2x = 725; method 2: let π 4 − x = α, then sin α = 35, so sin2x = sin (π 2 − 2 α) = Cos2 α = 1 − 2sin2 α = 725
The left and right focus of ellipse C x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 are F1 and F2 respectively. The point P (a, b) satisfies the absolute value of PF2 = the absolute value of F1F2. Calculate the eccentricity e of ellipse
Cos (x-pai / 4) = 3 / 5, then sin (x + Pai / 4)
Using induction formula
sin(x+π/4)
=sin[π/2-(π/4-x)]
=cos(π/4-x)
=cos(x-π/4)
=3/5
If the line y-kx-1 = 0 (K ∈ R) and the ellipse X25 + y2m = 1 have a common point, then the value range of M is & nbsp; ()
A. M ＞ 5B. 0 ＜ m ＜ 5C. M ＞ 1D. M ≥ 1 and m ≠ 5
The straight line y = KX + 1 passes through the point (0, 1), and the straight line y = KX + 1 and the ellipse have a common point. Therefore, when (0, 1) on the ellipse or in the ellipse ﹥ 0 + 1m ≤ 1 ﹥ m ≥ 1 and M = 25, the curve is a circle or not an ellipse, so m ≠ 25 the value range of real number m is: m ≥ 1 and m ≠ 25, so select D
If the line y = KX + 2 and the ellipse x + 4Y = 4 always have a common point, then the value range of the real number k is the same
According to the meaning of the problem: the system of equations: y = KX + 2 x & sup2; + 4Y & sup2; = 4 has a constant solution, that is: (4K & sup2; + 1) x & sup2; + 16kx + 12 = 0 △ = 0, k = ± √ 3 / 2, at this time, the line is tangent to the ellipse, and because the line y = KX + 2 passes through the point (0,2), so the
The line y = is substituted into the elliptic equation. The obtained x equation is simplified. When deata is less than or equal to 0. The range of K comes out

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