In the ellipse x ^ 2 / M + y ^ 2 / 4 = 1, the relationship between the semi major axis A and the semi focal length C is a = √ 2c, then M=

In the ellipse x ^ 2 / M + y ^ 2 / 4 = 1, the relationship between the semi major axis A and the semi focal length C is a = √ 2c, then M=

From the elliptic equation, we know that the semi major axis a = √ m, the semi minor axis B = √ 4 = 2, the semi focal length C = √ (a ^ 2-B ^ 2) = √ (M-4)
So a = √ M = √ 2C = √ (2m-8), so m = 2m-8, M = 8
It is known that the center of the ellipse and hyperbola with common focus is at the origin, the focus is on the x-axis, and the left and right focus are F1F2 respectively, and their focus in the first quadrant is p. the triangle pf1f2 is an isosceles triangle with Pf1 as the bottom. If the length of Pf1 is 10, and the value range of eccentricity of hyperbola (1,2), what is the value range of eccentricity of the ellipse?
Let the semi major axis length of ellipse be m (XM, YM), the semi real axis length of hyperbola be A2, C, | Pf1 | = m, | PF2 | = n, then {m + n = 2A1 M-N = 2A2 M = 10 N = 2C & # 8658; {A1 = 5 + Ca2 = 5-c, the problem is transformed into the known 1 ＜ C / 5-c ＜ 2, and the value range of C / 5 + C is obtained. Let C / 5-c = x
Question supplement: for hyperbola and ellipse with common focus, the center is the origin, the focus is on the x-axis, and the left and right focus, where a1 and A2 are a of hyperbola and ellipse respectively. Since E1 = cga1, the value range is (1,2), A1 will be
If cos (π 4 + x) = 35, the value of sin2x is ()
A. -2425B. -725C. 2425D. 725
From the known cos (π 4 + x) = 35, cos (π 2 + 2x) = 2cos2 (x + π 4) - 1 = 2 × 925-1 = - 725, i.e. - sin2x = - 725, sin2x = 725, so D
Judge whether the points a (- 2,7), B (5 / 3, - 42 / 5), C (1, - 14), D (2,7) are on the same inverse scale function image and which points are on the same function image, and find out the analytic expression of the function
The analytic expression of inverse proportion function is y = K / x, that is, xy = k, (k is a constant)
A（-2,7）,xy=-14
B(5/3,-42/5),xy=-14
C(1,-14),xy=-14
D(2,7),xy=14
So the three points a, B and C are on the same inverse scale function image, and the analytic expression of this function is xy = - 14, that is y = - 14 / X
It is known that the proposition p: the equation x22m-y2m − 1 = 1 represents the ellipse with the focus on the Y axis; the proposition q: the eccentricity e ∈ (1,2) of hyperbola y25-x2m = 1. If the propositions P and Q satisfy that P ∧ q is false and P ∨ q is true, the value range of M is obtained
From P: m − 1 ＜ 01 − m ＞ 2m2m ＞ 0 {0 ＜ m ＜ 13 (4 points) from proposition q: m ＞ 012 ＜ 5 + M5 ＜ 22 {0 ＜ m ＜ 15 (8 points) according to the known propositions P and Q: P ∧ q is false and P ∨ q is true, combined with two conditions, P is false and Q is true, so the value range of M is 13 ≤ m < 15 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp (12 points)
Given SiNx = √ 5-1 / 2, find the value of sin (x-pai / 4)
cosx=√(1-sinx^2)
sin(x-∏/4)=(√2/2)(sinx-cosx)
You can know by substitution~
The image of the inverse scale function passes through two points a (1 / A, 2 / a), B (2A / A-1, 1-A / a), and the analytic expression of the inverse scale function is obtained if the point C (m, 1) is on the function image
The image of the inverse scale function passes through two points a (1 / A, 2 / a), B (2A / A-1, - (1-A / a)) to find the analytic expression of the inverse scale function. If the point C (m, 1) is on the image of the function, the area of △ ABC can be found
The analytic expression of the inverse scale function is y = 2 / X. A (- 1, - 2) B (1,2). Since C is on the image of the inverse scale function, C (2,1). The line y = X-1 where AC is located intersects the x-axis at d (1,0), a and B cross the origin, and B, C intersects the x-axis at e (3,0). So s △ ABC = s △ OEB + s △ oad-s △ Dec = 3 + 1 -- 1 = 3
It is known that the hyperbola and the square of ellipse X / 9 + the square of ellipse Y / 25 = 1 are in common focus, and the sum of their eccentricities is 14 / 5
The equation of ellipse is X & sup2 / 9 + Y & sup2 / 25 = 1, a = 5, B = 3. C = 4E = C / A, e = 4 / 5, the eccentricity of hyperbola is 14 / 5-4 / 5 = 2, because the focus of hyperbola is C = 4, E = C / a = 4 / a = 2, so a = 2B & sup2; = 16-4 = 12, the focus is on Y axis, so the equation of hyperbola is Y & sup2 / 4-x & sup2 / 12 = 1
It is known that the hyperbola and the square of ellipse X / 9 + the square of ellipse Y / 25 = 1 are confocal
The focus of ellipse is (plus or minus 4,0), which is also the focus of hyperbola, then hyperbola C is 4
Next, the eccentricity of ellipse is 4 / 5, and the sum of eccentricity is 14 / 5, so the hyperbolic eccentricity is 2
So the hyperbola C / A is 2, and C is equal to 4, so a is equal to 2. Then the hyperbola formula can be obtained by calculating B
It is known that the hyperbola and the square of ellipse X / 9 + the square of ellipse Y / 25 = 1 are confocal
The focus of ellipse is (plus or minus 4,0), which is also the focus of hyperbola, then hyperbola C is 4
Next, the eccentricity of ellipse is 4 / 5, and the sum of eccentricity is 14 / 5, so the hyperbolic eccentricity is 2
So the hyperbola C / A is 2, and C is equal to 4, so a is equal to 2. Then the hyperbola formula can be obtained by calculating B
Given that sina and cosa are two of the equations 3x & sup2; - 2 χ + a = 0, then the value of a is
Using Veda's theorem to solve that the sum of two is equal to - B / A, and the product of two is equal to C / a (not AHA in it)
How to quickly draw the image of inverse scale function according to the analytic expression of inverse scale function? What is the relationship between the size of K and X and the image of function?
When k is less than 0 and K is less than 0 in the first quadrant, and K is less than 0 and K in the third quadrant, Y increases with the increase of X. therefore, you can draw a sketch according to the value of K. to draw a standard, you can only find some simple coordinates