If the line y = K (x + 1) + 1 and the ellipse x2 / 5 + Y2 / M = 1 always have a common point, and the focus of the ellipse is on the x-axis, then the value range of M is

If the line y = K (x + 1) + 1 and the ellipse x2 / 5 + Y2 / M = 1 always have a common point, and the focus of the ellipse is on the x-axis, then the value range of M is

The line passes through the point (- 1,1), so the point must be in the ellipse,
Therefore, 1 / 5 + 1 / mm > 0
5>m>5/4
Given that f (x) = 1 + log, the logarithm of X with 2 as the base, X belongs to [1,4]. If f (x) - 2m-3 > 0 is constant, the value range of M is obtained
Given f (x) = 1 + log, the logarithm of X with base 2
The logarithm X of log with 2 as the base x belongs to [1,4], and the minimum value is 0
The minimum value of F (x) is 1
If f (x) - 2m-3 > 0, it is true
Then f (x) > 2m + 3
So 2m + 3
If the eccentricity of the ellipse x ^ 2 / 2 + y ^ 2 / m with focus on the axis is 1 / 2, then the value of the real number m is?
The focus is on the x-axis
c^2=a^2-b^2=2-m
e=c/a
1/4=(2-m)/2
m=3/2
The focus is on the Y axis
e^2=c^2/a^2
1/4=(m-2)/m
m=4m-8
m=8/3
3 / 2 or 8 / 3
The focus is on the x-axis
c^2=a^2-b^2=2-m
e=c/a
1/4=(2-m)/2
m=3/2
The focus is on the Y axis
e^2=c^2/a^2
1/4=(m-2)/m
M = 4m-8, it's hard
m=8/3
Loga (1 / 2) > = 1 / 4, that is, the logarithm of the bottom 1 / 2 of log is not less than 1 / 4, and the value range of a is calculated?
lg(1/2)/lga≥ 1/4
When a > 1, there are:
lg(1/2)≥ lga/4
That is: A ^ (1 / 4) ≤ 1 / 2, a ≤ 1 / 16 has no solution
When 0
loga(1/2)〉=1/4
So 1 / 4 of A
If the equation KX + y = 3 represents an ellipse with focus on the x-axis, then the value range of the real number k is
x²/(3/k)+y²/3=1
Ellipse with focus on X axis
therefore
3/k>3
Zero
Two thirds of logo
Zero
If the line y-kx-1 = 0 and the ellipse x2 / 5 + Y2 / M = 1 have a common point, then the value range of M is
The answer is [1,5] ∪ [5, positive infinity)
&So if the minor axis of the ellipse is smaller than 1, which is the case in the figure, there may be no intersection between the line and the ellipse. & nbsp; in order to ensure that there is an intersection, the other axis must be greater than 1 & nbsp; so √ M & gt; 1 & nbsp; get M & gt; 1. But one thing to know is that the equation ellipse can not be a circle, So m is not equal to 5, so the value range of M is (1,5) and (5, positive infinity)
If loga (2 / 3) is less than 1, then the value range of a?
Please explain in detail
When loga (2 / 3) 1, loga (2 / 3)
If all x belongs to R, the line y = KX + 2 and the ellipse x ^ 2 / 9 + y ^ 2 / m ^ 2 = 1 have a common point, then the value range of positive real number m
The answer is [2,3] ∪ (3, + ∞)
Constant crossing point of straight line (0,2)
To ensure a constant intersection point, only (0,2) on or in the ellipse is needed
0/9+4/m^2≤1
m^2≥4
M ≤ - 2 or m ≥ 2
Because m is a positive number, m ≥ 2
Because it is an ellipse, m ^ 2 ≠ 9
m≠±3
Get the answer
If you draw a picture, you will find that if the point is inside the ellipse, it must have an intersection with the ellipse
Let f (x) = 2 ^ x-log a x (the logarithm of X with a as the base) (a > 0, a ≠ 1), there is f (x) on (0,1 / 2)
1. When a > 1, on (0,1 / 2)
2^x>0
loga(x)0
So f (x) = 2 ^ x-log a (x) > 0 is not true
2. When 0