If we know the intersection of the square of ellipse X / 4 + y and the square of hyperbola X-Y / 2 = 1, and F1F2 is the left and right focus of ellipse, we can find the cos angle FPF It's the focus of ellipses and hyperbolas

If we know the intersection of the square of ellipse X / 4 + y and the square of hyperbola X-Y / 2 = 1, and F1F2 is the left and right focus of ellipse, we can find the cos angle FPF It's the focus of ellipses and hyperbolas

Elliptic equation: x ^ 2 / 4 + y ^ 2 = 1, A1 = 2, B1 = 1, C1 = √ 3, F1 (- √ 3,0), F2 (√ 3,0); hyperbolic equation: x ^ 2-y ^ 2 / 2 = 1, A2 = 1, B2 = √ 2, C2 = √ 3, F1 (- √ 3,0), F2 (√ 3,0); two curves have the same focus, P is on two curves at the same time, according to the definition of ellipse, | Pf1 | + | PF2 | = 2A1 = 4
Is p an ellipse or a point on a hyperbola
If the ellipse x ^ 2 / M + y ^ 2 = 1 (M > 0) and the hyperbola x ^ 2 / n-y ^ 2 = 1 (n > 0) have the same focus F1F2, P is an intersection of the two curves; the area of triangle f1pf2
Let m ＞ n ＞ 1 x ^ 2 / M + y ^ 2 = 1 (M > 0) and x ^ 2 / n-y ^ 2 = 1 (n > 0) have the same focus. M-1 = n + 1. M-N = 2 | F1F2 | = 2 (m-1), Y & sup2 of the intersection of x ^ 2 / M + y ^ 2 = 1 and x ^ 2 / n-y ^ 2 = 1; = (m-n) / (M + n) triangle f1pf2 area = √ 2 (m-1) / √ (M + n) = √
The answer is 1. It's not convenient for me to write the process...
F1F2 is the left and right focus of the ellipse x ^ 2 / 4 + y ^ 2 = 1
If the sum of the moving points of the ellipse is the maximum of the left and right points of the ellipse
It is easy to know that a = 2, B = 1, C = root 3, so F1 (- root 3,0), F2 (root 3,0), let P (x, y), then vector Pf1 × vector PF2 = (- root 3-x, y) × (root 3-x, - y) = x ^ 2 + y ^ 2-3 = x ^ 2 + 1 - (x ^ 2 / 4) - 3 = (3x ^ 2-8) / 4. Because it belongs to [- 2,2], when x = 0, that is, P is the endpoint of ellipse minor axis, the minimum value of vector Pf1 × vector PF2 is
Definition of ellipse and hyperbola and definition of Quasilinear in high school mathematics
An ellipse defines the locus of a point on a plane whose sum of distances from one point to two fixed points is equal to a constant
Hyperbola is defined as the locus of a point on the plane whose absolute value of the distance difference between a point and two fixed points is equal to a constant
The ratio of the distance from a moving point to a fixed point and to a fixed line is equal to the constant E 0
The image of the first-order function y = (m-2) x + 3m-3 passes through the first, second and fourth quadrants to find the value range of M
∵ when the image of the first-order function y = (m-2) x + 3m-3 passes through the first, second and fourth quadrants, ∵ m − 2 ＜ 03m − 3 ＞ 0, the solution is 1 ＜ m ＜ 2
If the images of positive scale functions y = KX and y = 2x are symmetric about X axis, then the value of K is equal to_____ .
The point (1,2) passing through y = 2x is (1, - 2) symmetric about X axis, which is substituted into y = KX, k = - 2
But what I want to ask is why there is a point (1,2)?
k=-2
(1,2) is a point on the function y = 2x,
In fact, you can take any point on it. You can take (2,4) or (3,6)
The result is the same
Let 1 / (COS ^ 2 a-SiN ^ 2 a) be expressed as Tana
1/(cos^2 a-sin^2 a)
=（cos^2 a+sin^2 a）/（cos^2 a-sin^2 a）
Then divide by cos ^ 2
=(1+tan^2a)/(1-tan^2a)
If the image of the linear function y = (2-m) x + m passes through the first, second and fourth quadrants, then the value range of M is______ .
According to the meaning of the question, we get 2-m ＜ 0 and m ＞ 0, then we get m ＞ 2
For any real number k, the positional relationship between the circle C: X Λ 2 + y + 2-6x-8y + 12 = 0 and the line L: kx-y-4k + 3 = 0
The circle and square are changed into standard form
(x-3)^2+(y-4)^2=13
Center coordinates (3,4)
The distance from the center of a circle to a known straight line is obtained by the formula of the distance from a point to a straight line
d=|3k-4-4k+3|/√[k^2+(-1)^2]=|k+1|/√(k^2+1)
d^2=(k+1)^2/(k^2+1)
From the mean inequality
k^2+1≥2k
(k+1)^2≤2(k^2+1)
d^2≤2
Prove cos ^ 4 a-SiN ^ 4 a = cos ^ 2 (1-tana) (1 + Tana)
Left = (COS & sup2; a + Sin & sup2; a) (COS & sup2; a-SiN & sup2; a)
=1*cos2a
=cos2a
Right = cos & sup2; a (1-sina / COSA) (1 + Sina / COSA)
=cos²a(1-sin²a/cos²a)
=cos²a-sin²a
=cos2a
Left = right
Proof of proposition
Instead of Tana = Sina / cosa on the right,
That is: cos ^ 2 (1-tana) (1 + Tana)
=cos^2 a-sin^2 a
The formula is multiplied by 1
cos^2(1-tana)(1+tana)
=（cos^2 a-sin^2 a）（cos^2 a+sin^2 a）
=cos^4 a-sin^4 a
cos^4 a-sin^4 a
=(cos^2a-sin^2 a)(cos^2a+sin^2 a)
=cos^2a-sin^2 a
=cos2a
cos^2a(1-tana)(1+tana)
=cos^2a(1-tan^2a)
=cos^2a-cos^2atan^2a)
=cos^2a-sin^2 a
=cos2a
So cos ^ 4 a-SiN ^ 4 a = cos ^ 2A (1-tana) (1 + Tana)
cos^4 a-sin^4 a
=(cos^2a-sin^2 a)(cos^2a+sin^2 a)
=cos^2a-sin^2 a
=cos2a
cos^2a(1-tana)(1+tana)
=cos^2a(1-tan^2a)
=cos^2a-cos^2atan^2a)
=cos^2a-sin^2 a
=cos2a
cos^4 a-sin^4 a=cos^2a(1-tana)(1+tana)