When the value of the first-order function y = (4 + 2m) x + M-4 is known, the image passes through one three four quadrants

When the value of the first-order function y = (4 + 2m) x + M-4 is known, the image passes through one three four quadrants

∵ when y = (4 + 2m) x + M-4, the image passes through one three four quadrants (k > 0, B < 0)
∴4+2m﹥0,m-4﹤0
∴m﹥-2,m﹤4
When - 2 ＜ m ＜ 4 and the value of the first-order function y = (4 + 2m) x + M-4, the image passes through one three four quadrants
A function passes through one, three, four quadrants
The intersection of the slope greater than 0 and the Y axis is below the X axis
So y = (4 + 2m) x + M-4
4+2m>0 m>-2
M-4
Ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1, (a > b > 0) has a half focal length of C. if the point (C, 2C) is on the ellipse, then the eccentricity of the ellipse is e
If (C, 2C) is on the ellipse, then: C (C, 2C) points (C, 2C) on the ellipse, then: C (C, 2C) points (C, 2C) on the ellipse, then: C (C, 2C) points (C, 2C) on the ellipse, then: C (C, 2C) at the ellipse, then: C (C, 2C) at the ellipse, then: C (C 3535;#35\35\35\#3535;#########35#\\\\\\\\\\\\\｛ 178; C ｛ 178; = B ｛ 178; (a ｛ 178; - C ｛ 178;) 4A ｛ 178; C ｛ 178
In order to make the image of function y = (2m-3) x + (3N + 1) pass through the first, second and fourth quadrants, the value range of M and n should be
1,2,4, then the coefficient of X of the function is less than zero, then 2m-30, then the value range of n is n greater than one third of negative~~~
If the abscissa of the intersection point of the line y = 2x and the ellipse is exactly C, then the eccentricity of the ellipse is C______ .
From the meaning of the question, the ordinate of the intersection of the line y = 2x and the ellipse is 2C. Substituting it into x2a2 + y2b2 = 1, we get c2a2 + 4c2b2 = 1 and ∧ E2 + 4e21 − E2 = 1, so e = 2 − 1. The other root is not the meaning of the question, so the answer is 2 − 1
If the image of function y = (2m + 3) x + m + 1 does not pass through the second quadrant, then the value range of M is
Because the function doesn't go through the second quadrant
When 2m + 3 = 0, then M = - 1.5 and the function is y = - 0.5
When 2m + 3 > 0, then M + 1
The focal point of the oval billiard plate is a and B. the long axis is 2a and the focal length is 2C
Points a and B are its focal points. The length of the long axis is 2a and the focal length is 2C. When the small ball resting on point a (the radius of the small ball is ignored) starts from point a along a straight line and returns to point a for the first time after being reflected by the elliptical wall, the distance that the small ball passes through is ()?
A.4a
B.2(a-c)
C.2(a+c)
D. A.B. and C
(from the focus of the movement along the long axis of the situation is not counted ah) you Big Brother Big Sister help little brother
D. For any point: except for the two vertices of the major axis, the distance from any point on the ellipse to the two focal points is 2A. Moreover, starting from any focal point and reflecting by the surface of the ellipse, it must return to another focal point. Therefore, for any point, it must start from a, go to a point on the ellipse, then to B, then to a point on the ellipse, and then to a, so it is two groups of line segments connecting two focal points, so 4A pairs
For a special point, the closest vertex to a, just start from a, pass through the vertex, and then return to a, so the distance is only 2 (A-C), so B is right
For another special point, the vertex far from a, just start from a, arrive at the center, arrive at B, arrive at the vertex, and return. So the total length is 2 (a + C)
So ABC is right, choose D!
If the image of the linear function y = (3-m) x + m passes through the first, second and third quadrants, then the value range of M is
If the image of a linear function y = (3-m) x + m passes through the first, second and third quadrants,
Then, 3-m > 0, and, M > 0
Thus, 0 < m < 3
From the meaning of the title
3-m＞0
m＞0
therefore
0＜m＜3
3-m > 0, and M > 0
0＜m＜3
It is known that the square of X ＋ the square of Y ＋ the square of a is equal to 1. The distances from a point to two focal points are D1 and D2 respectively, and the focal length is 2C. If D1, 2C and D2 form an arithmetic sequence, then the eccentricity of the ellipse is?
∵ d1,2c, D2 into arithmetic sequence, ∵ D1 + D2 = 2 * 2C = 4C
According to the definition of ellipse: D1 + D2 = 2A (the sum of the distances from a point on the ellipse to two focal points is 2a)
∴2a=4c,∴e=c/a=1/2
That is, the centrifugal rate is 1 / 2
If the image of primary function y = 2x + B does not pass through the second quadrant, try to determine the value range of B []
First calculate the condition of the function in the second quadrant
Y=2X+B>0,X0
On the contrary, it is not in the second quadrant B
It is known that the eccentricity of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is 1 / 2, F1 and F2 are the left and right focus of ellipse C respectively, if the focal length of ellipse C is 2
(1) (2) let m be any point on the ellipse, take M as the center of the circle, MF1 as the radius, and make the circle M. when the circle m has a common point with the right guide line L of the ellipse, find the maximum area of △ mf1f2
(1) ∵ 2C = 2 and C / a = 1 / 2,
∴c=1,a=2．
∴b²=3．
∴x²／4+y²／3=1．
(2) Let m (x0, Y0),
x0²／4+y0²／3=1．
∵F1（-1,0）,a²／c=4,
The line LX = 4
Because m and l share a common point,
The distance between M and l 4-x0 is less than or equal to the radius r of the circle
R²=MF1²=（x0+1）²+y0²,
（4-x0）²≤（x0+1）2+y0²,
y0²+10x0-15≥0．
∵y0²=3(1-x0²／4),
3-3x0²／4+10x0-15≥0．
∴4／3≤x0≤2．
When x0 = 4 / 3, | Y0 | = √ 15 / 3,
(S△MF1F2)max=1／2×2×√15／3=√15／3．